Mixing
How do I solve for $a$ and $b$ in $limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right)...
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Given $limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 2$
I need to solve for $a$ and $b$, so here we go,
$limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right)$
$= limlimits_{x to ∞} x left(2 +(3+x) left(ln (1+frac{a}{x}) - ln(1+frac{b}{x}) right) right)$
$=limlimits_{x to ∞} x left(2 +(3+x)left( dfrac{a-b}{x} right) right)$
$=limlimits_{x to ∞} left(2x +(3+x)left( a-b right) right)$
$=limlimits_{x to ∞} left(2x + 3(a-b) + x(a-b) right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!
limits limits-without-lhopital
asked Jan 19 at 4:24
William William
1,207414
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add a comment |
$begingroup$
Given $limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 2$
I need to solve for $a$ and $b$, so here we go,
$limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right)$
$= limlimits_{x to ∞} x left(2 +(3+x) left(ln (1+frac{a}{x}) - ln(1+frac{b}{x}) right) right)$
$=limlimits_{x to ∞} x left(2 +(3+x)left( dfrac{a-b}{x} right) right)$
$=limlimits_{x to ∞} left(2x +(3+x)left( a-b right) right)$
$=limlimits_{x to ∞} left(2x + 3(a-b) + x(a-b) right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!
limits limits-without-lhopital
asked Jan 19 at 4:24
William William
1,207414
$endgroup$
$begingroup$
Do you know for a fact that a solution exists?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:37
add a comment |
$begingroup$
Given $limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 2$
I need to solve for $a$ and $b$, so here we go,
$limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right)$
$= limlimits_{x to ∞} x left(2 +(3+x) left(ln (1+frac{a}{x}) - ln(1+frac{b}{x}) right) right)$
$=limlimits_{x to ∞} x left(2 +(3+x)left( dfrac{a-b}{x} right) right)$
$=limlimits_{x to ∞} left(2x +(3+x)left( a-b right) right)$
$=limlimits_{x to ∞} left(2x + 3(a-b) + x(a-b) right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!
limits limits-without-lhopital
asked Jan 19 at 4:24
William William
1,207414
$endgroup$
Given $limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 2$
I need to solve for $a$ and $b$, so here we go,
$limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right)$
$= limlimits_{x to ∞} x left(2 +(3+x) left(ln (1+frac{a}{x}) - ln(1+frac{b}{x}) right) right)$
$=limlimits_{x to ∞} x left(2 +(3+x)left( dfrac{a-b}{x} right) right)$
$=limlimits_{x to ∞} left(2x +(3+x)left( a-b right) right)$
$=limlimits_{x to ∞} left(2x + 3(a-b) + x(a-b) right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!
limits limits-without-lhopital
limits limits-without-lhopital
asked Jan 19 at 4:24
William William
1,207414
asked Jan 19 at 4:24
William William
1,207414
asked Jan 19 at 4:24
William William
1,207414
asked Jan 19 at 4:24
William William
1,207414
asked Jan 19 at 4:24
William William
1,207414
1,207414
$begingroup$
Do you know for a fact that a solution exists?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:37
add a comment |
$begingroup$
Do you know for a fact that a solution exists?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:37
$begingroup$
Do you know for a fact that a solution exists?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:37
$begingroup$
Do you know for a fact that a solution exists?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:37
add a comment |
5 Answers
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oldest
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Your mistake lies in the stp where you replace the difference between logs with the ratio $(a-b) /x$. You can not replace an expression $A$ by another expression $B$ unless $A=B$. While evaluating limit you may find replacements of $A$ by $B$ even when $Aneq B$ but such replacements always represent some hidden steps which the the author assumes to be obvious (an example of such a replacement is in my solution where the factor $x$ is replaced by $x+b$ and it is based on the fact that $x/(x+b) to 1$ as $xtoinfty$). Also see this answer for details about such replacements.
The given limit condition implies that $(x+3)log((x+a)/(x+b))to - 2$ Then clearly $aneq b$ and we have $$(a-b)cdotfrac{x+3}{x+b}cdotdfrac{logleft(1+dfrac{a-b}{x+b}right)}{dfrac{a-b}{x+b}}to - 2$$ Thus $a-b=-2$ or $a=b-2$. Using this value of $a$ in original limit we see that $$xleft(2+(x+3)logleft(1-frac{2}{x+b}right)right)to 2$$ Replacing the first factor by $x+b$ and putting $x+b=2/t$ we see that $$lim_{tto 0^{+}}frac{1}{t}left(2+left(frac{2}{t}+3-bright)log(1-t)right)=1$$ or $$frac{2t+2log(1-t)}{t^2}+(3-b)frac{log(1-t)}{t}to 1$$ By L'Hospital's Rule or Taylor series the first fraction tends to $-1$ so that $-1+b-3=1$ or $b=5$ and $a=3$.
answered Jan 19 at 7:21
Paramanand SinghParamanand Singh
50.4k556166
$endgroup$
$begingroup$
Thank you for such a detailed answer, I didn't want the solution like the most answers here, I only wanted someone to point out mistakes in mine, like you did. I knew those rules, I was almost certain I could seperate the log expression from the rest of the expression. Guess I was quick to jump to the conclusions.
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– William
Jan 19 at 9:39
add a comment |
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Hint: you have to expand $log (1+y)$ up to the term in $y^{2}$ to answer this question. $log (1+y)=y-frac {y^{2}} 2+0(y^{2})$
answered Jan 19 at 4:57
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
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$begingroup$
Then my next obvious question would be why doesn't $limlimits_{xto 0} dfrac{ln ( 1+ x)}{x} = 1 $ work here? When $x to infty Rightarrow frac{a}{x} to 0$ in which case the identity should work right? All I've done is multiply and divide by $frac{a}{x}$ I know Maclaurin Series may work here but still why is this wrong?
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– William
Jan 19 at 5:52
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@William: well the limit $limlimits _{xto 0}dfrac{log (1+x)}{x}=1$ does work here (see my answer) but it does not mean that you can replace $log(1+x)$ with $x$ but rather it means that you can replace the expression $limlimits _{xto 0}dfrac{log(1+x)}{x}$ with $1$. See more details in beginning of my answer.
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– Paramanand Singh
Jan 19 at 8:51
add a comment |
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Hint
First of all, let $x=frac 1y$ to make the expression
$$A=frac{(3 y+1) log left(frac{1+a y}{1+b y}right)+2 y}{y^2}$$ Now, using Taylor series, you should have
$$A=frac{a-b+2}{y}+frac{1}{2} left(-a^2+6 a+b^2-6 bright)+frac{1}{6} y left(2
a^3-9 a^2-2 b^3+9 b^2right)+Oleft(y^2right)$$ So
$$a-b+2=0 qquad text{and} qquad frac{1}{2} left(-a^2+6 a+b^2-6 bright)=2$$ This is easy to solve.
answered Jan 19 at 5:59
Claude LeiboviciClaude Leibovici
123k1157134
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But can you tell me then what is wrong with my method?
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– William
Jan 19 at 6:12
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@William. Use one more term in the expansions.
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– Claude Leibovici
Jan 19 at 6:20
add a comment |
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Set $1/x=h$
$$lim_{hto0^+}dfrac{2h+(3h+1){ln(1+ah)-ln(1+bh)}}{h^2}$$
$$=3left(alim_{hto0^+}dfrac{ln(1+ah)-1}{ah}-blim_{hto0^+}dfrac{ln(1+bh)-1}{bh}right)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+a^2lim_{hto0^+}dfrac{ln(1+ah)-ah}{(ah)^2}-b^2lim_{hto0^+}dfrac{ln(1+bh)-bh}{(bh)^2}+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$
$$=3(a-b)+dfrac{b^2-a^2}2+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$ using Are all limits solvable without L'Hôpital Rule or Series Expansion,
Clearly, we need $2+a-b=0$ and $3(a-b)+dfrac{b^2-a^2}2=2$
answered Jan 19 at 6:17
lab bhattacharjeelab bhattacharjee
226k15157275
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Since the limit exists, $2x$ and $x(a−b)$ must cancel out so $a−b=2$.
You wanted to say $2+a-b=0 Rightarrow color{red}{a-b=-2}$.
And you cannot go from the difference of logarithms (line $2$) to the difference of arguments (line $3$), because it is not a factor, but a sum $(2+cdots)$, which is multiplied by $x$. In other words, you can not apply the asymptotics to an addend, but must apply to a factor.
Alternatively:
$$limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 2 Rightarrow \
limlimits_{x to ∞} frac{2 +(3+x) ln left( dfrac{x+a}{x+b} right)}{frac 1x} = 2 (*)$$
For the limit to exist:
$$limlimits_{x to ∞} left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 0 Rightarrow cdots Rightarrow a-b=-2.$$
Now use L'Hospital's rule for $(*)$ twice:
$$limlimits_{x to ∞} frac{ln (x+a) +frac{3+x}{x+a}- ln (x+b) -frac{3+x}{x+b}}{-frac 1{x^2}} = \
=limlimits_{x to ∞} frac{frac1{x+a}+frac{a-3}{(x+a)^2}-frac1{x+b}-frac{b-3}{(x+b)^2}}{frac 2{x^3}} = \
=lim_{xtoinfty} -frac{x^3 (a - b) (2 a b + a x - 3 a + b x - 3 b - 6 x)}{2 (x+a)^2 (x+b)^2}=2 Rightarrow \
color{red}{a+b-6=2}.$$
Hence:
$$begin{cases}a-b=-2 \ a+b-6=2 end{cases} Rightarrow a=3, b=5.$$
answered Jan 19 at 7:12
farruhotafarruhota
20.5k2740
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5 Answers
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5 Answers
5
active
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$begingroup$
Your mistake lies in the stp where you replace the difference between logs with the ratio $(a-b) /x$. You can not replace an expression $A$ by another expression $B$ unless $A=B$. While evaluating limit you may find replacements of $A$ by $B$ even when $Aneq B$ but such replacements always represent some hidden steps which the the author assumes to be obvious (an example of such a replacement is in my solution where the factor $x$ is replaced by $x+b$ and it is based on the fact that $x/(x+b) to 1$ as $xtoinfty$). Also see this answer for details about such replacements.
The given limit condition implies that $(x+3)log((x+a)/(x+b))to - 2$ Then clearly $aneq b$ and we have $$(a-b)cdotfrac{x+3}{x+b}cdotdfrac{logleft(1+dfrac{a-b}{x+b}right)}{dfrac{a-b}{x+b}}to - 2$$ Thus $a-b=-2$ or $a=b-2$. Using this value of $a$ in original limit we see that $$xleft(2+(x+3)logleft(1-frac{2}{x+b}right)right)to 2$$ Replacing the first factor by $x+b$ and putting $x+b=2/t$ we see that $$lim_{tto 0^{+}}frac{1}{t}left(2+left(frac{2}{t}+3-bright)log(1-t)right)=1$$ or $$frac{2t+2log(1-t)}{t^2}+(3-b)frac{log(1-t)}{t}to 1$$ By L'Hospital's Rule or Taylor series the first fraction tends to $-1$ so that $-1+b-3=1$ or $b=5$ and $a=3$.
answered Jan 19 at 7:21
Paramanand SinghParamanand Singh
50.4k556166
$endgroup$
$begingroup$
Thank you for such a detailed answer, I didn't want the solution like the most answers here, I only wanted someone to point out mistakes in mine, like you did. I knew those rules, I was almost certain I could seperate the log expression from the rest of the expression. Guess I was quick to jump to the conclusions.
$endgroup$
– William
Jan 19 at 9:39
add a comment |
$begingroup$
Your mistake lies in the stp where you replace the difference between logs with the ratio $(a-b) /x$. You can not replace an expression $A$ by another expression $B$ unless $A=B$. While evaluating limit you may find replacements of $A$ by $B$ even when $Aneq B$ but such replacements always represent some hidden steps which the the author assumes to be obvious (an example of such a replacement is in my solution where the factor $x$ is replaced by $x+b$ and it is based on the fact that $x/(x+b) to 1$ as $xtoinfty$). Also see this answer for details about such replacements.
The given limit condition implies that $(x+3)log((x+a)/(x+b))to - 2$ Then clearly $aneq b$ and we have $$(a-b)cdotfrac{x+3}{x+b}cdotdfrac{logleft(1+dfrac{a-b}{x+b}right)}{dfrac{a-b}{x+b}}to - 2$$ Thus $a-b=-2$ or $a=b-2$. Using this value of $a$ in original limit we see that $$xleft(2+(x+3)logleft(1-frac{2}{x+b}right)right)to 2$$ Replacing the first factor by $x+b$ and putting $x+b=2/t$ we see that $$lim_{tto 0^{+}}frac{1}{t}left(2+left(frac{2}{t}+3-bright)log(1-t)right)=1$$ or $$frac{2t+2log(1-t)}{t^2}+(3-b)frac{log(1-t)}{t}to 1$$ By L'Hospital's Rule or Taylor series the first fraction tends to $-1$ so that $-1+b-3=1$ or $b=5$ and $a=3$.
answered Jan 19 at 7:21
Paramanand SinghParamanand Singh
50.4k556166
$endgroup$
$begingroup$
Thank you for such a detailed answer, I didn't want the solution like the most answers here, I only wanted someone to point out mistakes in mine, like you did. I knew those rules, I was almost certain I could seperate the log expression from the rest of the expression. Guess I was quick to jump to the conclusions.
$endgroup$
– William
Jan 19 at 9:39
add a comment |
$begingroup$
Your mistake lies in the stp where you replace the difference between logs with the ratio $(a-b) /x$. You can not replace an expression $A$ by another expression $B$ unless $A=B$. While evaluating limit you may find replacements of $A$ by $B$ even when $Aneq B$ but such replacements always represent some hidden steps which the the author assumes to be obvious (an example of such a replacement is in my solution where the factor $x$ is replaced by $x+b$ and it is based on the fact that $x/(x+b) to 1$ as $xtoinfty$). Also see this answer for details about such replacements.
The given limit condition implies that $(x+3)log((x+a)/(x+b))to - 2$ Then clearly $aneq b$ and we have $$(a-b)cdotfrac{x+3}{x+b}cdotdfrac{logleft(1+dfrac{a-b}{x+b}right)}{dfrac{a-b}{x+b}}to - 2$$ Thus $a-b=-2$ or $a=b-2$. Using this value of $a$ in original limit we see that $$xleft(2+(x+3)logleft(1-frac{2}{x+b}right)right)to 2$$ Replacing the first factor by $x+b$ and putting $x+b=2/t$ we see that $$lim_{tto 0^{+}}frac{1}{t}left(2+left(frac{2}{t}+3-bright)log(1-t)right)=1$$ or $$frac{2t+2log(1-t)}{t^2}+(3-b)frac{log(1-t)}{t}to 1$$ By L'Hospital's Rule or Taylor series the first fraction tends to $-1$ so that $-1+b-3=1$ or $b=5$ and $a=3$.
answered Jan 19 at 7:21
Paramanand SinghParamanand Singh
50.4k556166
$endgroup$
Your mistake lies in the stp where you replace the difference between logs with the ratio $(a-b) /x$. You can not replace an expression $A$ by another expression $B$ unless $A=B$. While evaluating limit you may find replacements of $A$ by $B$ even when $Aneq B$ but such replacements always represent some hidden steps which the the author assumes to be obvious (an example of such a replacement is in my solution where the factor $x$ is replaced by $x+b$ and it is based on the fact that $x/(x+b) to 1$ as $xtoinfty$). Also see this answer for details about such replacements.
The given limit condition implies that $(x+3)log((x+a)/(x+b))to - 2$ Then clearly $aneq b$ and we have $$(a-b)cdotfrac{x+3}{x+b}cdotdfrac{logleft(1+dfrac{a-b}{x+b}right)}{dfrac{a-b}{x+b}}to - 2$$ Thus $a-b=-2$ or $a=b-2$. Using this value of $a$ in original limit we see that $$xleft(2+(x+3)logleft(1-frac{2}{x+b}right)right)to 2$$ Replacing the first factor by $x+b$ and putting $x+b=2/t$ we see that $$lim_{tto 0^{+}}frac{1}{t}left(2+left(frac{2}{t}+3-bright)log(1-t)right)=1$$ or $$frac{2t+2log(1-t)}{t^2}+(3-b)frac{log(1-t)}{t}to 1$$ By L'Hospital's Rule or Taylor series the first fraction tends to $-1$ so that $-1+b-3=1$ or $b=5$ and $a=3$.
answered Jan 19 at 7:21
Paramanand SinghParamanand Singh
50.4k556166
edited Jan 19 at 8:45
answered Jan 19 at 7:21
Paramanand SinghParamanand Singh
50.4k556166
answered Jan 19 at 7:21
Paramanand SinghParamanand Singh
50.4k556166
answered Jan 19 at 7:21
Paramanand SinghParamanand Singh
50.4k556166
50.4k556166
$begingroup$
Thank you for such a detailed answer, I didn't want the solution like the most answers here, I only wanted someone to point out mistakes in mine, like you did. I knew those rules, I was almost certain I could seperate the log expression from the rest of the expression. Guess I was quick to jump to the conclusions.
$endgroup$
– William
Jan 19 at 9:39
add a comment |
$begingroup$
Thank you for such a detailed answer, I didn't want the solution like the most answers here, I only wanted someone to point out mistakes in mine, like you did. I knew those rules, I was almost certain I could seperate the log expression from the rest of the expression. Guess I was quick to jump to the conclusions.
$endgroup$
– William
Jan 19 at 9:39
$begingroup$
Thank you for such a detailed answer, I didn't want the solution like the most answers here, I only wanted someone to point out mistakes in mine, like you did. I knew those rules, I was almost certain I could seperate the log expression from the rest of the expression. Guess I was quick to jump to the conclusions.
$endgroup$
– William
Jan 19 at 9:39
$begingroup$
Thank you for such a detailed answer, I didn't want the solution like the most answers here, I only wanted someone to point out mistakes in mine, like you did. I knew those rules, I was almost certain I could seperate the log expression from the rest of the expression. Guess I was quick to jump to the conclusions.
$endgroup$
– William
Jan 19 at 9:39
add a comment |
$begingroup$
Hint: you have to expand $log (1+y)$ up to the term in $y^{2}$ to answer this question. $log (1+y)=y-frac {y^{2}} 2+0(y^{2})$
answered Jan 19 at 4:57
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
$begingroup$
Then my next obvious question would be why doesn't $limlimits_{xto 0} dfrac{ln ( 1+ x)}{x} = 1 $ work here? When $x to infty Rightarrow frac{a}{x} to 0$ in which case the identity should work right? All I've done is multiply and divide by $frac{a}{x}$ I know Maclaurin Series may work here but still why is this wrong?
$endgroup$
– William
Jan 19 at 5:52
$begingroup$
@William: well the limit $limlimits _{xto 0}dfrac{log (1+x)}{x}=1$ does work here (see my answer) but it does not mean that you can replace $log(1+x)$ with $x$ but rather it means that you can replace the expression $limlimits _{xto 0}dfrac{log(1+x)}{x}$ with $1$. See more details in beginning of my answer.
$endgroup$
– Paramanand Singh
Jan 19 at 8:51
add a comment |
$begingroup$
Hint: you have to expand $log (1+y)$ up to the term in $y^{2}$ to answer this question. $log (1+y)=y-frac {y^{2}} 2+0(y^{2})$
answered Jan 19 at 4:57
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
$begingroup$
Then my next obvious question would be why doesn't $limlimits_{xto 0} dfrac{ln ( 1+ x)}{x} = 1 $ work here? When $x to infty Rightarrow frac{a}{x} to 0$ in which case the identity should work right? All I've done is multiply and divide by $frac{a}{x}$ I know Maclaurin Series may work here but still why is this wrong?
$endgroup$
– William
Jan 19 at 5:52
$begingroup$
@William: well the limit $limlimits _{xto 0}dfrac{log (1+x)}{x}=1$ does work here (see my answer) but it does not mean that you can replace $log(1+x)$ with $x$ but rather it means that you can replace the expression $limlimits _{xto 0}dfrac{log(1+x)}{x}$ with $1$. See more details in beginning of my answer.
$endgroup$
– Paramanand Singh
Jan 19 at 8:51
add a comment |
$begingroup$
Hint: you have to expand $log (1+y)$ up to the term in $y^{2}$ to answer this question. $log (1+y)=y-frac {y^{2}} 2+0(y^{2})$
answered Jan 19 at 4:57
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
$endgroup$
Hint: you have to expand $log (1+y)$ up to the term in $y^{2}$ to answer this question. $log (1+y)=y-frac {y^{2}} 2+0(y^{2})$
answered Jan 19 at 4:57
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 19 at 4:57
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 19 at 4:57
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
answered Jan 19 at 4:57
Kavi Rama MurthyKavi Rama Murthy
63.7k42463
63.7k42463
$begingroup$
Then my next obvious question would be why doesn't $limlimits_{xto 0} dfrac{ln ( 1+ x)}{x} = 1 $ work here? When $x to infty Rightarrow frac{a}{x} to 0$ in which case the identity should work right? All I've done is multiply and divide by $frac{a}{x}$ I know Maclaurin Series may work here but still why is this wrong?
$endgroup$
– William
Jan 19 at 5:52
$begingroup$
@William: well the limit $limlimits _{xto 0}dfrac{log (1+x)}{x}=1$ does work here (see my answer) but it does not mean that you can replace $log(1+x)$ with $x$ but rather it means that you can replace the expression $limlimits _{xto 0}dfrac{log(1+x)}{x}$ with $1$. See more details in beginning of my answer.
$endgroup$
– Paramanand Singh
Jan 19 at 8:51
add a comment |
$begingroup$
Then my next obvious question would be why doesn't $limlimits_{xto 0} dfrac{ln ( 1+ x)}{x} = 1 $ work here? When $x to infty Rightarrow frac{a}{x} to 0$ in which case the identity should work right? All I've done is multiply and divide by $frac{a}{x}$ I know Maclaurin Series may work here but still why is this wrong?
$endgroup$
– William
Jan 19 at 5:52
$begingroup$
@William: well the limit $limlimits _{xto 0}dfrac{log (1+x)}{x}=1$ does work here (see my answer) but it does not mean that you can replace $log(1+x)$ with $x$ but rather it means that you can replace the expression $limlimits _{xto 0}dfrac{log(1+x)}{x}$ with $1$. See more details in beginning of my answer.
$endgroup$
– Paramanand Singh
Jan 19 at 8:51
$begingroup$
Then my next obvious question would be why doesn't $limlimits_{xto 0} dfrac{ln ( 1+ x)}{x} = 1 $ work here? When $x to infty Rightarrow frac{a}{x} to 0$ in which case the identity should work right? All I've done is multiply and divide by $frac{a}{x}$ I know Maclaurin Series may work here but still why is this wrong?
$endgroup$
– William
Jan 19 at 5:52
$begingroup$
Then my next obvious question would be why doesn't $limlimits_{xto 0} dfrac{ln ( 1+ x)}{x} = 1 $ work here? When $x to infty Rightarrow frac{a}{x} to 0$ in which case the identity should work right? All I've done is multiply and divide by $frac{a}{x}$ I know Maclaurin Series may work here but still why is this wrong?
$endgroup$
– William
Jan 19 at 5:52
$begingroup$
@William: well the limit $limlimits _{xto 0}dfrac{log (1+x)}{x}=1$ does work here (see my answer) but it does not mean that you can replace $log(1+x)$ with $x$ but rather it means that you can replace the expression $limlimits _{xto 0}dfrac{log(1+x)}{x}$ with $1$. See more details in beginning of my answer.
$endgroup$
– Paramanand Singh
Jan 19 at 8:51
$begingroup$
@William: well the limit $limlimits _{xto 0}dfrac{log (1+x)}{x}=1$ does work here (see my answer) but it does not mean that you can replace $log(1+x)$ with $x$ but rather it means that you can replace the expression $limlimits _{xto 0}dfrac{log(1+x)}{x}$ with $1$. See more details in beginning of my answer.
$endgroup$
– Paramanand Singh
Jan 19 at 8:51
add a comment |
$begingroup$
Hint
First of all, let $x=frac 1y$ to make the expression
$$A=frac{(3 y+1) log left(frac{1+a y}{1+b y}right)+2 y}{y^2}$$ Now, using Taylor series, you should have
$$A=frac{a-b+2}{y}+frac{1}{2} left(-a^2+6 a+b^2-6 bright)+frac{1}{6} y left(2
a^3-9 a^2-2 b^3+9 b^2right)+Oleft(y^2right)$$ So
$$a-b+2=0 qquad text{and} qquad frac{1}{2} left(-a^2+6 a+b^2-6 bright)=2$$ This is easy to solve.
answered Jan 19 at 5:59
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$begingroup$
But can you tell me then what is wrong with my method?
$endgroup$
– William
Jan 19 at 6:12
$begingroup$
@William. Use one more term in the expansions.
$endgroup$
– Claude Leibovici
Jan 19 at 6:20
add a comment |
$begingroup$
Hint
First of all, let $x=frac 1y$ to make the expression
$$A=frac{(3 y+1) log left(frac{1+a y}{1+b y}right)+2 y}{y^2}$$ Now, using Taylor series, you should have
$$A=frac{a-b+2}{y}+frac{1}{2} left(-a^2+6 a+b^2-6 bright)+frac{1}{6} y left(2
a^3-9 a^2-2 b^3+9 b^2right)+Oleft(y^2right)$$ So
$$a-b+2=0 qquad text{and} qquad frac{1}{2} left(-a^2+6 a+b^2-6 bright)=2$$ This is easy to solve.
answered Jan 19 at 5:59
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
$begingroup$
But can you tell me then what is wrong with my method?
$endgroup$
– William
Jan 19 at 6:12
$begingroup$
@William. Use one more term in the expansions.
$endgroup$
– Claude Leibovici
Jan 19 at 6:20
add a comment |
$begingroup$
Hint
First of all, let $x=frac 1y$ to make the expression
$$A=frac{(3 y+1) log left(frac{1+a y}{1+b y}right)+2 y}{y^2}$$ Now, using Taylor series, you should have
$$A=frac{a-b+2}{y}+frac{1}{2} left(-a^2+6 a+b^2-6 bright)+frac{1}{6} y left(2
a^3-9 a^2-2 b^3+9 b^2right)+Oleft(y^2right)$$ So
$$a-b+2=0 qquad text{and} qquad frac{1}{2} left(-a^2+6 a+b^2-6 bright)=2$$ This is easy to solve.
answered Jan 19 at 5:59
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
Hint
First of all, let $x=frac 1y$ to make the expression
$$A=frac{(3 y+1) log left(frac{1+a y}{1+b y}right)+2 y}{y^2}$$ Now, using Taylor series, you should have
$$A=frac{a-b+2}{y}+frac{1}{2} left(-a^2+6 a+b^2-6 bright)+frac{1}{6} y left(2
a^3-9 a^2-2 b^3+9 b^2right)+Oleft(y^2right)$$ So
$$a-b+2=0 qquad text{and} qquad frac{1}{2} left(-a^2+6 a+b^2-6 bright)=2$$ This is easy to solve.
answered Jan 19 at 5:59
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 5:59
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 5:59
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 5:59
Claude LeiboviciClaude Leibovici
123k1157134
123k1157134
$begingroup$
But can you tell me then what is wrong with my method?
$endgroup$
– William
Jan 19 at 6:12
$begingroup$
@William. Use one more term in the expansions.
$endgroup$
– Claude Leibovici
Jan 19 at 6:20
add a comment |
$begingroup$
But can you tell me then what is wrong with my method?
$endgroup$
– William
Jan 19 at 6:12
$begingroup$
@William. Use one more term in the expansions.
$endgroup$
– Claude Leibovici
Jan 19 at 6:20
$begingroup$
But can you tell me then what is wrong with my method?
$endgroup$
– William
Jan 19 at 6:12
$begingroup$
But can you tell me then what is wrong with my method?
$endgroup$
– William
Jan 19 at 6:12
$begingroup$
@William. Use one more term in the expansions.
$endgroup$
– Claude Leibovici
Jan 19 at 6:20
$begingroup$
@William. Use one more term in the expansions.
$endgroup$
– Claude Leibovici
Jan 19 at 6:20
add a comment |
$begingroup$
Set $1/x=h$
$$lim_{hto0^+}dfrac{2h+(3h+1){ln(1+ah)-ln(1+bh)}}{h^2}$$
$$=3left(alim_{hto0^+}dfrac{ln(1+ah)-1}{ah}-blim_{hto0^+}dfrac{ln(1+bh)-1}{bh}right)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+a^2lim_{hto0^+}dfrac{ln(1+ah)-ah}{(ah)^2}-b^2lim_{hto0^+}dfrac{ln(1+bh)-bh}{(bh)^2}+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$
$$=3(a-b)+dfrac{b^2-a^2}2+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$ using Are all limits solvable without L'Hôpital Rule or Series Expansion,
Clearly, we need $2+a-b=0$ and $3(a-b)+dfrac{b^2-a^2}2=2$
answered Jan 19 at 6:17
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Set $1/x=h$
$$lim_{hto0^+}dfrac{2h+(3h+1){ln(1+ah)-ln(1+bh)}}{h^2}$$
$$=3left(alim_{hto0^+}dfrac{ln(1+ah)-1}{ah}-blim_{hto0^+}dfrac{ln(1+bh)-1}{bh}right)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+a^2lim_{hto0^+}dfrac{ln(1+ah)-ah}{(ah)^2}-b^2lim_{hto0^+}dfrac{ln(1+bh)-bh}{(bh)^2}+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$
$$=3(a-b)+dfrac{b^2-a^2}2+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$ using Are all limits solvable without L'Hôpital Rule or Series Expansion,
Clearly, we need $2+a-b=0$ and $3(a-b)+dfrac{b^2-a^2}2=2$
answered Jan 19 at 6:17
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Set $1/x=h$
$$lim_{hto0^+}dfrac{2h+(3h+1){ln(1+ah)-ln(1+bh)}}{h^2}$$
$$=3left(alim_{hto0^+}dfrac{ln(1+ah)-1}{ah}-blim_{hto0^+}dfrac{ln(1+bh)-1}{bh}right)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+a^2lim_{hto0^+}dfrac{ln(1+ah)-ah}{(ah)^2}-b^2lim_{hto0^+}dfrac{ln(1+bh)-bh}{(bh)^2}+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$
$$=3(a-b)+dfrac{b^2-a^2}2+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$ using Are all limits solvable without L'Hôpital Rule or Series Expansion,
Clearly, we need $2+a-b=0$ and $3(a-b)+dfrac{b^2-a^2}2=2$
answered Jan 19 at 6:17
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Set $1/x=h$
$$lim_{hto0^+}dfrac{2h+(3h+1){ln(1+ah)-ln(1+bh)}}{h^2}$$
$$=3left(alim_{hto0^+}dfrac{ln(1+ah)-1}{ah}-blim_{hto0^+}dfrac{ln(1+bh)-1}{bh}right)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+lim_{hto0^+}dfrac{2h+ln(1+ah)-ln(1+bh)}{h^2}$$
$$=3(a-b)+a^2lim_{hto0^+}dfrac{ln(1+ah)-ah}{(ah)^2}-b^2lim_{hto0^+}dfrac{ln(1+bh)-bh}{(bh)^2}+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$
$$=3(a-b)+dfrac{b^2-a^2}2+lim_{hto0^+}dfrac{(2+a-b)h}{h^2}$$ using Are all limits solvable without L'Hôpital Rule or Series Expansion,
Clearly, we need $2+a-b=0$ and $3(a-b)+dfrac{b^2-a^2}2=2$
answered Jan 19 at 6:17
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 19 at 6:17
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 19 at 6:17
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 19 at 6:17
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
$begingroup$
Since the limit exists, $2x$ and $x(a−b)$ must cancel out so $a−b=2$.
You wanted to say $2+a-b=0 Rightarrow color{red}{a-b=-2}$.
And you cannot go from the difference of logarithms (line $2$) to the difference of arguments (line $3$), because it is not a factor, but a sum $(2+cdots)$, which is multiplied by $x$. In other words, you can not apply the asymptotics to an addend, but must apply to a factor.
Alternatively:
$$limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 2 Rightarrow \
limlimits_{x to ∞} frac{2 +(3+x) ln left( dfrac{x+a}{x+b} right)}{frac 1x} = 2 (*)$$
For the limit to exist:
$$limlimits_{x to ∞} left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 0 Rightarrow cdots Rightarrow a-b=-2.$$
Now use L'Hospital's rule for $(*)$ twice:
$$limlimits_{x to ∞} frac{ln (x+a) +frac{3+x}{x+a}- ln (x+b) -frac{3+x}{x+b}}{-frac 1{x^2}} = \
=limlimits_{x to ∞} frac{frac1{x+a}+frac{a-3}{(x+a)^2}-frac1{x+b}-frac{b-3}{(x+b)^2}}{frac 2{x^3}} = \
=lim_{xtoinfty} -frac{x^3 (a - b) (2 a b + a x - 3 a + b x - 3 b - 6 x)}{2 (x+a)^2 (x+b)^2}=2 Rightarrow \
color{red}{a+b-6=2}.$$
Hence:
$$begin{cases}a-b=-2 \ a+b-6=2 end{cases} Rightarrow a=3, b=5.$$
answered Jan 19 at 7:12
farruhotafarruhota
20.5k2740
$endgroup$
add a comment |
$begingroup$
Since the limit exists, $2x$ and $x(a−b)$ must cancel out so $a−b=2$.
You wanted to say $2+a-b=0 Rightarrow color{red}{a-b=-2}$.
And you cannot go from the difference of logarithms (line $2$) to the difference of arguments (line $3$), because it is not a factor, but a sum $(2+cdots)$, which is multiplied by $x$. In other words, you can not apply the asymptotics to an addend, but must apply to a factor.
Alternatively:
$$limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 2 Rightarrow \
limlimits_{x to ∞} frac{2 +(3+x) ln left( dfrac{x+a}{x+b} right)}{frac 1x} = 2 (*)$$
For the limit to exist:
$$limlimits_{x to ∞} left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 0 Rightarrow cdots Rightarrow a-b=-2.$$
Now use L'Hospital's rule for $(*)$ twice:
$$limlimits_{x to ∞} frac{ln (x+a) +frac{3+x}{x+a}- ln (x+b) -frac{3+x}{x+b}}{-frac 1{x^2}} = \
=limlimits_{x to ∞} frac{frac1{x+a}+frac{a-3}{(x+a)^2}-frac1{x+b}-frac{b-3}{(x+b)^2}}{frac 2{x^3}} = \
=lim_{xtoinfty} -frac{x^3 (a - b) (2 a b + a x - 3 a + b x - 3 b - 6 x)}{2 (x+a)^2 (x+b)^2}=2 Rightarrow \
color{red}{a+b-6=2}.$$
Hence:
$$begin{cases}a-b=-2 \ a+b-6=2 end{cases} Rightarrow a=3, b=5.$$
answered Jan 19 at 7:12
farruhotafarruhota
20.5k2740
$endgroup$
add a comment |
$begingroup$
Since the limit exists, $2x$ and $x(a−b)$ must cancel out so $a−b=2$.
You wanted to say $2+a-b=0 Rightarrow color{red}{a-b=-2}$.
And you cannot go from the difference of logarithms (line $2$) to the difference of arguments (line $3$), because it is not a factor, but a sum $(2+cdots)$, which is multiplied by $x$. In other words, you can not apply the asymptotics to an addend, but must apply to a factor.
Alternatively:
$$limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 2 Rightarrow \
limlimits_{x to ∞} frac{2 +(3+x) ln left( dfrac{x+a}{x+b} right)}{frac 1x} = 2 (*)$$
For the limit to exist:
$$limlimits_{x to ∞} left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 0 Rightarrow cdots Rightarrow a-b=-2.$$
Now use L'Hospital's rule for $(*)$ twice:
$$limlimits_{x to ∞} frac{ln (x+a) +frac{3+x}{x+a}- ln (x+b) -frac{3+x}{x+b}}{-frac 1{x^2}} = \
=limlimits_{x to ∞} frac{frac1{x+a}+frac{a-3}{(x+a)^2}-frac1{x+b}-frac{b-3}{(x+b)^2}}{frac 2{x^3}} = \
=lim_{xtoinfty} -frac{x^3 (a - b) (2 a b + a x - 3 a + b x - 3 b - 6 x)}{2 (x+a)^2 (x+b)^2}=2 Rightarrow \
color{red}{a+b-6=2}.$$
Hence:
$$begin{cases}a-b=-2 \ a+b-6=2 end{cases} Rightarrow a=3, b=5.$$
answered Jan 19 at 7:12
farruhotafarruhota
20.5k2740
$endgroup$
Since the limit exists, $2x$ and $x(a−b)$ must cancel out so $a−b=2$.
You wanted to say $2+a-b=0 Rightarrow color{red}{a-b=-2}$.
And you cannot go from the difference of logarithms (line $2$) to the difference of arguments (line $3$), because it is not a factor, but a sum $(2+cdots)$, which is multiplied by $x$. In other words, you can not apply the asymptotics to an addend, but must apply to a factor.
Alternatively:
$$limlimits_{x to ∞} x left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 2 Rightarrow \
limlimits_{x to ∞} frac{2 +(3+x) ln left( dfrac{x+a}{x+b} right)}{frac 1x} = 2 (*)$$
For the limit to exist:
$$limlimits_{x to ∞} left(2 +(3+x) ln left( dfrac{x+a}{x+b} right) right) = 0 Rightarrow cdots Rightarrow a-b=-2.$$
Now use L'Hospital's rule for $(*)$ twice:
$$limlimits_{x to ∞} frac{ln (x+a) +frac{3+x}{x+a}- ln (x+b) -frac{3+x}{x+b}}{-frac 1{x^2}} = \
=limlimits_{x to ∞} frac{frac1{x+a}+frac{a-3}{(x+a)^2}-frac1{x+b}-frac{b-3}{(x+b)^2}}{frac 2{x^3}} = \
=lim_{xtoinfty} -frac{x^3 (a - b) (2 a b + a x - 3 a + b x - 3 b - 6 x)}{2 (x+a)^2 (x+b)^2}=2 Rightarrow \
color{red}{a+b-6=2}.$$
Hence:
$$begin{cases}a-b=-2 \ a+b-6=2 end{cases} Rightarrow a=3, b=5.$$
answered Jan 19 at 7:12
farruhotafarruhota
20.5k2740
answered Jan 19 at 7:12
farruhotafarruhota
20.5k2740
answered Jan 19 at 7:12
farruhotafarruhota
20.5k2740
answered Jan 19 at 7:12
farruhotafarruhota
20.5k2740
20.5k2740
add a comment |
add a comment |
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$begingroup$
Do you know for a fact that a solution exists?
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:37