Mixing
Evaluating $lim_{xto0^+}frac{2x(sin...
$begingroup$
To start with, the term $xe^{-{1over x}}$ can be ignored.Then splitting the term $ln(frac{1+x^2}{1-x^2})=ln(1+x^2)-ln(1-x^2)$ and expanding both of them up to order 3 we have in the denominator $2x^2+{2x^6over 3}-2x^2={2x^6over3}$
As for the numerator $arctan(2x^3)sim2x^3$ and ${2x^7+x^8over 3x^2+x^4}sim{2over3}x^5$.So if we expand $sin x$ up to order 2,then take the square (ignoring higher order terms) and multiply by $2x$ we have $2x(sin x)^2sim2x^3-{2over3}x^5$ which cancels out with the other terms,giving us an indeterminate form.So I think we should have $x^6$ term in the numerator.
But if I expand $sin x$ up to order 3 and take square, I fail to obtain $x^5$ term(so that when multiplied by $2x$ gives me $x^6$).How can this be done?
calculus limits limits-without-lhopital
asked Jan 19 at 19:46
Turan NəsibliTuran Nəsibli
776
$endgroup$
|
show 2 more comments
$begingroup$
To start with, the term $xe^{-{1over x}}$ can be ignored.Then splitting the term $ln(frac{1+x^2}{1-x^2})=ln(1+x^2)-ln(1-x^2)$ and expanding both of them up to order 3 we have in the denominator $2x^2+{2x^6over 3}-2x^2={2x^6over3}$
As for the numerator $arctan(2x^3)sim2x^3$ and ${2x^7+x^8over 3x^2+x^4}sim{2over3}x^5$.So if we expand $sin x$ up to order 2,then take the square (ignoring higher order terms) and multiply by $2x$ we have $2x(sin x)^2sim2x^3-{2over3}x^5$ which cancels out with the other terms,giving us an indeterminate form.So I think we should have $x^6$ term in the numerator.
But if I expand $sin x$ up to order 3 and take square, I fail to obtain $x^5$ term(so that when multiplied by $2x$ gives me $x^6$).How can this be done?
calculus limits limits-without-lhopital
asked Jan 19 at 19:46
Turan NəsibliTuran Nəsibli
776
$endgroup$
$begingroup$
Where did you find this problem
$endgroup$
– Zach
Jan 19 at 19:57
1
$begingroup$
While I know you want to determine the steps to find the limit, that limit (according to Mathematica and a simple plot) is $1/2$.
$endgroup$
– JimB
Jan 19 at 19:58
$begingroup$
@Zach This is question from a previous exam of my course
$endgroup$
– Turan Nəsibli
Jan 19 at 19:58
$begingroup$
You really do need to evaluate each piece up to $x^6.$ There is cancellation among the lower degree terms, including some coefficients that come out $0$ in the first place.
$endgroup$
– Will Jagy
Jan 19 at 20:22
$begingroup$
It looks like $0$: desmos.com/calculator/nrq5rm6ooj
$endgroup$
– clathratus
Jan 19 at 20:44
|
show 2 more comments
$begingroup$
To start with, the term $xe^{-{1over x}}$ can be ignored.Then splitting the term $ln(frac{1+x^2}{1-x^2})=ln(1+x^2)-ln(1-x^2)$ and expanding both of them up to order 3 we have in the denominator $2x^2+{2x^6over 3}-2x^2={2x^6over3}$
As for the numerator $arctan(2x^3)sim2x^3$ and ${2x^7+x^8over 3x^2+x^4}sim{2over3}x^5$.So if we expand $sin x$ up to order 2,then take the square (ignoring higher order terms) and multiply by $2x$ we have $2x(sin x)^2sim2x^3-{2over3}x^5$ which cancels out with the other terms,giving us an indeterminate form.So I think we should have $x^6$ term in the numerator.
But if I expand $sin x$ up to order 3 and take square, I fail to obtain $x^5$ term(so that when multiplied by $2x$ gives me $x^6$).How can this be done?
calculus limits limits-without-lhopital
asked Jan 19 at 19:46
Turan NəsibliTuran Nəsibli
776
$endgroup$
To start with, the term $xe^{-{1over x}}$ can be ignored.Then splitting the term $ln(frac{1+x^2}{1-x^2})=ln(1+x^2)-ln(1-x^2)$ and expanding both of them up to order 3 we have in the denominator $2x^2+{2x^6over 3}-2x^2={2x^6over3}$
As for the numerator $arctan(2x^3)sim2x^3$ and ${2x^7+x^8over 3x^2+x^4}sim{2over3}x^5$.So if we expand $sin x$ up to order 2,then take the square (ignoring higher order terms) and multiply by $2x$ we have $2x(sin x)^2sim2x^3-{2over3}x^5$ which cancels out with the other terms,giving us an indeterminate form.So I think we should have $x^6$ term in the numerator.
But if I expand $sin x$ up to order 3 and take square, I fail to obtain $x^5$ term(so that when multiplied by $2x$ gives me $x^6$).How can this be done?
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
asked Jan 19 at 19:46
Turan NəsibliTuran Nəsibli
776
asked Jan 19 at 19:46
Turan NəsibliTuran Nəsibli
776
asked Jan 19 at 19:46
Turan NəsibliTuran Nəsibli
776
asked Jan 19 at 19:46
Turan NəsibliTuran Nəsibli
776
asked Jan 19 at 19:46
Turan NəsibliTuran Nəsibli
776
776
$begingroup$
Where did you find this problem
$endgroup$
– Zach
Jan 19 at 19:57
1
$begingroup$
While I know you want to determine the steps to find the limit, that limit (according to Mathematica and a simple plot) is $1/2$.
$endgroup$
– JimB
Jan 19 at 19:58
$begingroup$
@Zach This is question from a previous exam of my course
$endgroup$
– Turan Nəsibli
Jan 19 at 19:58
$begingroup$
You really do need to evaluate each piece up to $x^6.$ There is cancellation among the lower degree terms, including some coefficients that come out $0$ in the first place.
$endgroup$
– Will Jagy
Jan 19 at 20:22
$begingroup$
It looks like $0$: desmos.com/calculator/nrq5rm6ooj
$endgroup$
– clathratus
Jan 19 at 20:44
|
show 2 more comments
$begingroup$
Where did you find this problem
$endgroup$
– Zach
Jan 19 at 19:57
1
$begingroup$
While I know you want to determine the steps to find the limit, that limit (according to Mathematica and a simple plot) is $1/2$.
$endgroup$
– JimB
Jan 19 at 19:58
$begingroup$
@Zach This is question from a previous exam of my course
$endgroup$
– Turan Nəsibli
Jan 19 at 19:58
$begingroup$
You really do need to evaluate each piece up to $x^6.$ There is cancellation among the lower degree terms, including some coefficients that come out $0$ in the first place.
$endgroup$
– Will Jagy
Jan 19 at 20:22
$begingroup$
It looks like $0$: desmos.com/calculator/nrq5rm6ooj
$endgroup$
– clathratus
Jan 19 at 20:44
$begingroup$
Where did you find this problem
$endgroup$
– Zach
Jan 19 at 19:57
$begingroup$
Where did you find this problem
$endgroup$
– Zach
Jan 19 at 19:57
1
1
$begingroup$
While I know you want to determine the steps to find the limit, that limit (according to Mathematica and a simple plot) is $1/2$.
$endgroup$
– JimB
Jan 19 at 19:58
$begingroup$
While I know you want to determine the steps to find the limit, that limit (according to Mathematica and a simple plot) is $1/2$.
$endgroup$
– JimB
Jan 19 at 19:58
$begingroup$
@Zach This is question from a previous exam of my course
$endgroup$
– Turan Nəsibli
Jan 19 at 19:58
$begingroup$
@Zach This is question from a previous exam of my course
$endgroup$
– Turan Nəsibli
Jan 19 at 19:58
$begingroup$
You really do need to evaluate each piece up to $x^6.$ There is cancellation among the lower degree terms, including some coefficients that come out $0$ in the first place.
$endgroup$
– Will Jagy
Jan 19 at 20:22
$begingroup$
You really do need to evaluate each piece up to $x^6.$ There is cancellation among the lower degree terms, including some coefficients that come out $0$ in the first place.
$endgroup$
– Will Jagy
Jan 19 at 20:22
$begingroup$
It looks like $0$: desmos.com/calculator/nrq5rm6ooj
$endgroup$
– clathratus
Jan 19 at 20:44
$begingroup$
It looks like $0$: desmos.com/calculator/nrq5rm6ooj
$endgroup$
– clathratus
Jan 19 at 20:44
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
The items below are called Taylor Series. This is a good way to reduce errors, in cases where L'Hospital's method leads to messy derivatives. One may take time and get each series correct. In this case, all the series derive from that for $$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots, $$ which leads fairly quickly to series for $1/(3+t),$ for $log(1+t),$ and for $arctan t$
$$ $$
To confirm the first step below, multiply the right hand side by $x^2 + 3$ and notice the cancellations:
$$ frac{1}{x^2 + 3} = frac{1}{3} - frac{x^2}{9} + frac{x^4}{27} - frac{x^6}{81} + cdots $$
$$ frac{1}{x^4 + 3x^2} = frac{1}{3x^2} - frac{1}{9} + frac{x^2}{27} - frac{x^4}{81} + cdots $$
$$ frac{x^8 + 2x^7}{x^4 + 3x^2} = frac{2x^5}{3} + frac{x^6}{3} - frac{2x^7}{9} - frac{x^8}{9} + cdots $$
$$ 2x sin^2 x = 2 x^3 - frac{2x^5}{3} + frac{4x^7}{45} - cdots $$
$$ arctan (2x^3) = 2 x^3 - frac{8x^9}{3} - cdots $$
The numerator is $$ frac{x^6}{3} - frac{2x^7}{15} + cdots $$
The denominator is $$ frac{2x^6}{3} + frac{2x^{10}}{5} + cdots $$
giving limit $1/2$
answered Jan 19 at 20:47
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
How can we directly obtain Taylor series of $frac{1}{x^2+3}$ from $frac{1}{1+t}$?
$endgroup$
– Turan Nəsibli
Jan 19 at 22:55
1
$begingroup$
@TuranNəsibli substitute $t = frac{x^2}{3}.$ That gives the series for $frac{3}{3+x^2},$ so you need to divide the whole thing by $3$ when you are otherwise done.
$endgroup$
– Will Jagy
Jan 19 at 23:13
add a comment |
$begingroup$
You made a mistake by considering only the first asymptotic term of the algebraic fraction. Be careful:
$$frac{2x^5+x^6}{3+x^2}=frac{2}{3}x^5 + frac{1}{3}x^6 + o(x^6).$$
Then your limit gives $frac{1}{2}$, as results from the comment by JimB.
answered Jan 19 at 20:37
MatteoMatteo
838311
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I didn't really get the way you obtained it,could you give me more details?
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– Turan Nəsibli
Jan 19 at 20:46
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@TuranNəsibli $$frac{2x^5+x^6}{3+x^2} = frac{2}{3}x^5 +left(-frac{2}{3}x^5 + frac{2x^5+x^6}{3+x^2}right)$$ and then look what is left in parenthesis.
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– Matteo
Jan 19 at 20:56
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I don't recall seeing anything like this throughout course.While simplifying the fraction I just neglected higher order terms both in numerator and denominator and obtained ${2over3}x^5$
$endgroup$
– Turan Nəsibli
Jan 19 at 21:03
$begingroup$
@TuranNəsibli this only gives you dominant term, which cancels out with other terms in the overall numerator. In that case you must consider also what you had neglected before, as when you further develop the MacLaurin polynomial of sine or logarithm. By the way of course you'd obtain the same results taking derivatives, or as shown in another answer. I usually proceed as I showed you here.
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– Matteo
Jan 19 at 21:08
1
$begingroup$
Exacly, you can use @WillJagy approach, which is very fine, of course. Or, as I showed you right above, here, by subracting dominant term to the original fraction.
$endgroup$
– Matteo
Jan 19 at 21:17
|
show 7 more comments
$begingroup$
The Taylor series approach suggested by @Matteo and @WillJagy is the way to go. Here is that approach made a bit more explicit:
- Divide the numerator and denominator by x^6.
- Construct the resulting first order Taylor series for both numerator and denominator.
- Observe what the limits must be.
$$lim_{xto 0^+} , frac{-tan ^{-1}left(2 x^3right)+frac{(x+2) x^5}{x^2+3}+2 x sin ^2(x)}{x exp left(-frac{1}{x}right)-2 x^2+log left(frac{x^2+1}{1-x^2}right)}$$
$$lim_{xto 0^+} ,{ {(frac{1}{3}-frac{2 x}{15}+o(x^2) )}over{(frac{2}{3}+e^{-1/x}(frac{1}{x^5}+o(x^2)))}}=frac{1}{2}$$
Here is a plot of the function:
answered Jan 19 at 23:38
JimBJimB
55537
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The items below are called Taylor Series. This is a good way to reduce errors, in cases where L'Hospital's method leads to messy derivatives. One may take time and get each series correct. In this case, all the series derive from that for $$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots, $$ which leads fairly quickly to series for $1/(3+t),$ for $log(1+t),$ and for $arctan t$
$$ $$
To confirm the first step below, multiply the right hand side by $x^2 + 3$ and notice the cancellations:
$$ frac{1}{x^2 + 3} = frac{1}{3} - frac{x^2}{9} + frac{x^4}{27} - frac{x^6}{81} + cdots $$
$$ frac{1}{x^4 + 3x^2} = frac{1}{3x^2} - frac{1}{9} + frac{x^2}{27} - frac{x^4}{81} + cdots $$
$$ frac{x^8 + 2x^7}{x^4 + 3x^2} = frac{2x^5}{3} + frac{x^6}{3} - frac{2x^7}{9} - frac{x^8}{9} + cdots $$
$$ 2x sin^2 x = 2 x^3 - frac{2x^5}{3} + frac{4x^7}{45} - cdots $$
$$ arctan (2x^3) = 2 x^3 - frac{8x^9}{3} - cdots $$
The numerator is $$ frac{x^6}{3} - frac{2x^7}{15} + cdots $$
The denominator is $$ frac{2x^6}{3} + frac{2x^{10}}{5} + cdots $$
giving limit $1/2$
answered Jan 19 at 20:47
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
How can we directly obtain Taylor series of $frac{1}{x^2+3}$ from $frac{1}{1+t}$?
$endgroup$
– Turan Nəsibli
Jan 19 at 22:55
1
$begingroup$
@TuranNəsibli substitute $t = frac{x^2}{3}.$ That gives the series for $frac{3}{3+x^2},$ so you need to divide the whole thing by $3$ when you are otherwise done.
$endgroup$
– Will Jagy
Jan 19 at 23:13
add a comment |
$begingroup$
The items below are called Taylor Series. This is a good way to reduce errors, in cases where L'Hospital's method leads to messy derivatives. One may take time and get each series correct. In this case, all the series derive from that for $$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots, $$ which leads fairly quickly to series for $1/(3+t),$ for $log(1+t),$ and for $arctan t$
$$ $$
To confirm the first step below, multiply the right hand side by $x^2 + 3$ and notice the cancellations:
$$ frac{1}{x^2 + 3} = frac{1}{3} - frac{x^2}{9} + frac{x^4}{27} - frac{x^6}{81} + cdots $$
$$ frac{1}{x^4 + 3x^2} = frac{1}{3x^2} - frac{1}{9} + frac{x^2}{27} - frac{x^4}{81} + cdots $$
$$ frac{x^8 + 2x^7}{x^4 + 3x^2} = frac{2x^5}{3} + frac{x^6}{3} - frac{2x^7}{9} - frac{x^8}{9} + cdots $$
$$ 2x sin^2 x = 2 x^3 - frac{2x^5}{3} + frac{4x^7}{45} - cdots $$
$$ arctan (2x^3) = 2 x^3 - frac{8x^9}{3} - cdots $$
The numerator is $$ frac{x^6}{3} - frac{2x^7}{15} + cdots $$
The denominator is $$ frac{2x^6}{3} + frac{2x^{10}}{5} + cdots $$
giving limit $1/2$
answered Jan 19 at 20:47
Will JagyWill Jagy
104k5102201
$endgroup$
$begingroup$
How can we directly obtain Taylor series of $frac{1}{x^2+3}$ from $frac{1}{1+t}$?
$endgroup$
– Turan Nəsibli
Jan 19 at 22:55
1
$begingroup$
@TuranNəsibli substitute $t = frac{x^2}{3}.$ That gives the series for $frac{3}{3+x^2},$ so you need to divide the whole thing by $3$ when you are otherwise done.
$endgroup$
– Will Jagy
Jan 19 at 23:13
add a comment |
$begingroup$
The items below are called Taylor Series. This is a good way to reduce errors, in cases where L'Hospital's method leads to messy derivatives. One may take time and get each series correct. In this case, all the series derive from that for $$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots, $$ which leads fairly quickly to series for $1/(3+t),$ for $log(1+t),$ and for $arctan t$
$$ $$
To confirm the first step below, multiply the right hand side by $x^2 + 3$ and notice the cancellations:
$$ frac{1}{x^2 + 3} = frac{1}{3} - frac{x^2}{9} + frac{x^4}{27} - frac{x^6}{81} + cdots $$
$$ frac{1}{x^4 + 3x^2} = frac{1}{3x^2} - frac{1}{9} + frac{x^2}{27} - frac{x^4}{81} + cdots $$
$$ frac{x^8 + 2x^7}{x^4 + 3x^2} = frac{2x^5}{3} + frac{x^6}{3} - frac{2x^7}{9} - frac{x^8}{9} + cdots $$
$$ 2x sin^2 x = 2 x^3 - frac{2x^5}{3} + frac{4x^7}{45} - cdots $$
$$ arctan (2x^3) = 2 x^3 - frac{8x^9}{3} - cdots $$
The numerator is $$ frac{x^6}{3} - frac{2x^7}{15} + cdots $$
The denominator is $$ frac{2x^6}{3} + frac{2x^{10}}{5} + cdots $$
giving limit $1/2$
answered Jan 19 at 20:47
Will JagyWill Jagy
104k5102201
$endgroup$
The items below are called Taylor Series. This is a good way to reduce errors, in cases where L'Hospital's method leads to messy derivatives. One may take time and get each series correct. In this case, all the series derive from that for $$ frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 cdots, $$ which leads fairly quickly to series for $1/(3+t),$ for $log(1+t),$ and for $arctan t$
$$ $$
To confirm the first step below, multiply the right hand side by $x^2 + 3$ and notice the cancellations:
$$ frac{1}{x^2 + 3} = frac{1}{3} - frac{x^2}{9} + frac{x^4}{27} - frac{x^6}{81} + cdots $$
$$ frac{1}{x^4 + 3x^2} = frac{1}{3x^2} - frac{1}{9} + frac{x^2}{27} - frac{x^4}{81} + cdots $$
$$ frac{x^8 + 2x^7}{x^4 + 3x^2} = frac{2x^5}{3} + frac{x^6}{3} - frac{2x^7}{9} - frac{x^8}{9} + cdots $$
$$ 2x sin^2 x = 2 x^3 - frac{2x^5}{3} + frac{4x^7}{45} - cdots $$
$$ arctan (2x^3) = 2 x^3 - frac{8x^9}{3} - cdots $$
The numerator is $$ frac{x^6}{3} - frac{2x^7}{15} + cdots $$
The denominator is $$ frac{2x^6}{3} + frac{2x^{10}}{5} + cdots $$
giving limit $1/2$
answered Jan 19 at 20:47
Will JagyWill Jagy
104k5102201
edited Jan 19 at 21:31
answered Jan 19 at 20:47
Will JagyWill Jagy
104k5102201
answered Jan 19 at 20:47
Will JagyWill Jagy
104k5102201
answered Jan 19 at 20:47
Will JagyWill Jagy
104k5102201
104k5102201
$begingroup$
How can we directly obtain Taylor series of $frac{1}{x^2+3}$ from $frac{1}{1+t}$?
$endgroup$
– Turan Nəsibli
Jan 19 at 22:55
1
$begingroup$
@TuranNəsibli substitute $t = frac{x^2}{3}.$ That gives the series for $frac{3}{3+x^2},$ so you need to divide the whole thing by $3$ when you are otherwise done.
$endgroup$
– Will Jagy
Jan 19 at 23:13
add a comment |
$begingroup$
How can we directly obtain Taylor series of $frac{1}{x^2+3}$ from $frac{1}{1+t}$?
$endgroup$
– Turan Nəsibli
Jan 19 at 22:55
1
$begingroup$
@TuranNəsibli substitute $t = frac{x^2}{3}.$ That gives the series for $frac{3}{3+x^2},$ so you need to divide the whole thing by $3$ when you are otherwise done.
$endgroup$
– Will Jagy
Jan 19 at 23:13
$begingroup$
How can we directly obtain Taylor series of $frac{1}{x^2+3}$ from $frac{1}{1+t}$?
$endgroup$
– Turan Nəsibli
Jan 19 at 22:55
$begingroup$
How can we directly obtain Taylor series of $frac{1}{x^2+3}$ from $frac{1}{1+t}$?
$endgroup$
– Turan Nəsibli
Jan 19 at 22:55
1
1
$begingroup$
@TuranNəsibli substitute $t = frac{x^2}{3}.$ That gives the series for $frac{3}{3+x^2},$ so you need to divide the whole thing by $3$ when you are otherwise done.
$endgroup$
– Will Jagy
Jan 19 at 23:13
$begingroup$
@TuranNəsibli substitute $t = frac{x^2}{3}.$ That gives the series for $frac{3}{3+x^2},$ so you need to divide the whole thing by $3$ when you are otherwise done.
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– Will Jagy
Jan 19 at 23:13
add a comment |
$begingroup$
You made a mistake by considering only the first asymptotic term of the algebraic fraction. Be careful:
$$frac{2x^5+x^6}{3+x^2}=frac{2}{3}x^5 + frac{1}{3}x^6 + o(x^6).$$
Then your limit gives $frac{1}{2}$, as results from the comment by JimB.
answered Jan 19 at 20:37
MatteoMatteo
838311
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$begingroup$
I didn't really get the way you obtained it,could you give me more details?
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– Turan Nəsibli
Jan 19 at 20:46
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@TuranNəsibli $$frac{2x^5+x^6}{3+x^2} = frac{2}{3}x^5 +left(-frac{2}{3}x^5 + frac{2x^5+x^6}{3+x^2}right)$$ and then look what is left in parenthesis.
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– Matteo
Jan 19 at 20:56
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I don't recall seeing anything like this throughout course.While simplifying the fraction I just neglected higher order terms both in numerator and denominator and obtained ${2over3}x^5$
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– Turan Nəsibli
Jan 19 at 21:03
$begingroup$
@TuranNəsibli this only gives you dominant term, which cancels out with other terms in the overall numerator. In that case you must consider also what you had neglected before, as when you further develop the MacLaurin polynomial of sine or logarithm. By the way of course you'd obtain the same results taking derivatives, or as shown in another answer. I usually proceed as I showed you here.
$endgroup$
– Matteo
Jan 19 at 21:08
1
$begingroup$
Exacly, you can use @WillJagy approach, which is very fine, of course. Or, as I showed you right above, here, by subracting dominant term to the original fraction.
$endgroup$
– Matteo
Jan 19 at 21:17
|
show 7 more comments
$begingroup$
You made a mistake by considering only the first asymptotic term of the algebraic fraction. Be careful:
$$frac{2x^5+x^6}{3+x^2}=frac{2}{3}x^5 + frac{1}{3}x^6 + o(x^6).$$
Then your limit gives $frac{1}{2}$, as results from the comment by JimB.
answered Jan 19 at 20:37
MatteoMatteo
838311
$endgroup$
$begingroup$
I didn't really get the way you obtained it,could you give me more details?
$endgroup$
– Turan Nəsibli
Jan 19 at 20:46
$begingroup$
@TuranNəsibli $$frac{2x^5+x^6}{3+x^2} = frac{2}{3}x^5 +left(-frac{2}{3}x^5 + frac{2x^5+x^6}{3+x^2}right)$$ and then look what is left in parenthesis.
$endgroup$
– Matteo
Jan 19 at 20:56
$begingroup$
I don't recall seeing anything like this throughout course.While simplifying the fraction I just neglected higher order terms both in numerator and denominator and obtained ${2over3}x^5$
$endgroup$
– Turan Nəsibli
Jan 19 at 21:03
$begingroup$
@TuranNəsibli this only gives you dominant term, which cancels out with other terms in the overall numerator. In that case you must consider also what you had neglected before, as when you further develop the MacLaurin polynomial of sine or logarithm. By the way of course you'd obtain the same results taking derivatives, or as shown in another answer. I usually proceed as I showed you here.
$endgroup$
– Matteo
Jan 19 at 21:08
1
$begingroup$
Exacly, you can use @WillJagy approach, which is very fine, of course. Or, as I showed you right above, here, by subracting dominant term to the original fraction.
$endgroup$
– Matteo
Jan 19 at 21:17
|
show 7 more comments
$begingroup$
You made a mistake by considering only the first asymptotic term of the algebraic fraction. Be careful:
$$frac{2x^5+x^6}{3+x^2}=frac{2}{3}x^5 + frac{1}{3}x^6 + o(x^6).$$
Then your limit gives $frac{1}{2}$, as results from the comment by JimB.
answered Jan 19 at 20:37
MatteoMatteo
838311
$endgroup$
You made a mistake by considering only the first asymptotic term of the algebraic fraction. Be careful:
$$frac{2x^5+x^6}{3+x^2}=frac{2}{3}x^5 + frac{1}{3}x^6 + o(x^6).$$
Then your limit gives $frac{1}{2}$, as results from the comment by JimB.
answered Jan 19 at 20:37
MatteoMatteo
838311
edited Jan 20 at 3:59
answered Jan 19 at 20:37
MatteoMatteo
838311
answered Jan 19 at 20:37
MatteoMatteo
838311
answered Jan 19 at 20:37
MatteoMatteo
838311
838311
$begingroup$
I didn't really get the way you obtained it,could you give me more details?
$endgroup$
– Turan Nəsibli
Jan 19 at 20:46
$begingroup$
@TuranNəsibli $$frac{2x^5+x^6}{3+x^2} = frac{2}{3}x^5 +left(-frac{2}{3}x^5 + frac{2x^5+x^6}{3+x^2}right)$$ and then look what is left in parenthesis.
$endgroup$
– Matteo
Jan 19 at 20:56
$begingroup$
I don't recall seeing anything like this throughout course.While simplifying the fraction I just neglected higher order terms both in numerator and denominator and obtained ${2over3}x^5$
$endgroup$
– Turan Nəsibli
Jan 19 at 21:03
$begingroup$
@TuranNəsibli this only gives you dominant term, which cancels out with other terms in the overall numerator. In that case you must consider also what you had neglected before, as when you further develop the MacLaurin polynomial of sine or logarithm. By the way of course you'd obtain the same results taking derivatives, or as shown in another answer. I usually proceed as I showed you here.
$endgroup$
– Matteo
Jan 19 at 21:08
1
$begingroup$
Exacly, you can use @WillJagy approach, which is very fine, of course. Or, as I showed you right above, here, by subracting dominant term to the original fraction.
$endgroup$
– Matteo
Jan 19 at 21:17
|
show 7 more comments
$begingroup$
I didn't really get the way you obtained it,could you give me more details?
$endgroup$
– Turan Nəsibli
Jan 19 at 20:46
$begingroup$
@TuranNəsibli $$frac{2x^5+x^6}{3+x^2} = frac{2}{3}x^5 +left(-frac{2}{3}x^5 + frac{2x^5+x^6}{3+x^2}right)$$ and then look what is left in parenthesis.
$endgroup$
– Matteo
Jan 19 at 20:56
$begingroup$
I don't recall seeing anything like this throughout course.While simplifying the fraction I just neglected higher order terms both in numerator and denominator and obtained ${2over3}x^5$
$endgroup$
– Turan Nəsibli
Jan 19 at 21:03
$begingroup$
@TuranNəsibli this only gives you dominant term, which cancels out with other terms in the overall numerator. In that case you must consider also what you had neglected before, as when you further develop the MacLaurin polynomial of sine or logarithm. By the way of course you'd obtain the same results taking derivatives, or as shown in another answer. I usually proceed as I showed you here.
$endgroup$
– Matteo
Jan 19 at 21:08
1
$begingroup$
Exacly, you can use @WillJagy approach, which is very fine, of course. Or, as I showed you right above, here, by subracting dominant term to the original fraction.
$endgroup$
– Matteo
Jan 19 at 21:17
$begingroup$
I didn't really get the way you obtained it,could you give me more details?
$endgroup$
– Turan Nəsibli
Jan 19 at 20:46
$begingroup$
I didn't really get the way you obtained it,could you give me more details?
$endgroup$
– Turan Nəsibli
Jan 19 at 20:46
$begingroup$
@TuranNəsibli $$frac{2x^5+x^6}{3+x^2} = frac{2}{3}x^5 +left(-frac{2}{3}x^5 + frac{2x^5+x^6}{3+x^2}right)$$ and then look what is left in parenthesis.
$endgroup$
– Matteo
Jan 19 at 20:56
$begingroup$
@TuranNəsibli $$frac{2x^5+x^6}{3+x^2} = frac{2}{3}x^5 +left(-frac{2}{3}x^5 + frac{2x^5+x^6}{3+x^2}right)$$ and then look what is left in parenthesis.
$endgroup$
– Matteo
Jan 19 at 20:56
$begingroup$
I don't recall seeing anything like this throughout course.While simplifying the fraction I just neglected higher order terms both in numerator and denominator and obtained ${2over3}x^5$
$endgroup$
– Turan Nəsibli
Jan 19 at 21:03
$begingroup$
I don't recall seeing anything like this throughout course.While simplifying the fraction I just neglected higher order terms both in numerator and denominator and obtained ${2over3}x^5$
$endgroup$
– Turan Nəsibli
Jan 19 at 21:03
$begingroup$
@TuranNəsibli this only gives you dominant term, which cancels out with other terms in the overall numerator. In that case you must consider also what you had neglected before, as when you further develop the MacLaurin polynomial of sine or logarithm. By the way of course you'd obtain the same results taking derivatives, or as shown in another answer. I usually proceed as I showed you here.
$endgroup$
– Matteo
Jan 19 at 21:08
$begingroup$
@TuranNəsibli this only gives you dominant term, which cancels out with other terms in the overall numerator. In that case you must consider also what you had neglected before, as when you further develop the MacLaurin polynomial of sine or logarithm. By the way of course you'd obtain the same results taking derivatives, or as shown in another answer. I usually proceed as I showed you here.
$endgroup$
– Matteo
Jan 19 at 21:08
1
1
$begingroup$
Exacly, you can use @WillJagy approach, which is very fine, of course. Or, as I showed you right above, here, by subracting dominant term to the original fraction.
$endgroup$
– Matteo
Jan 19 at 21:17
$begingroup$
Exacly, you can use @WillJagy approach, which is very fine, of course. Or, as I showed you right above, here, by subracting dominant term to the original fraction.
$endgroup$
– Matteo
Jan 19 at 21:17
|
show 7 more comments
$begingroup$
The Taylor series approach suggested by @Matteo and @WillJagy is the way to go. Here is that approach made a bit more explicit:
- Divide the numerator and denominator by x^6.
- Construct the resulting first order Taylor series for both numerator and denominator.
- Observe what the limits must be.
$$lim_{xto 0^+} , frac{-tan ^{-1}left(2 x^3right)+frac{(x+2) x^5}{x^2+3}+2 x sin ^2(x)}{x exp left(-frac{1}{x}right)-2 x^2+log left(frac{x^2+1}{1-x^2}right)}$$
$$lim_{xto 0^+} ,{ {(frac{1}{3}-frac{2 x}{15}+o(x^2) )}over{(frac{2}{3}+e^{-1/x}(frac{1}{x^5}+o(x^2)))}}=frac{1}{2}$$
Here is a plot of the function:
answered Jan 19 at 23:38
JimBJimB
55537
$endgroup$
add a comment |
$begingroup$
The Taylor series approach suggested by @Matteo and @WillJagy is the way to go. Here is that approach made a bit more explicit:
- Divide the numerator and denominator by x^6.
- Construct the resulting first order Taylor series for both numerator and denominator.
- Observe what the limits must be.
$$lim_{xto 0^+} , frac{-tan ^{-1}left(2 x^3right)+frac{(x+2) x^5}{x^2+3}+2 x sin ^2(x)}{x exp left(-frac{1}{x}right)-2 x^2+log left(frac{x^2+1}{1-x^2}right)}$$
$$lim_{xto 0^+} ,{ {(frac{1}{3}-frac{2 x}{15}+o(x^2) )}over{(frac{2}{3}+e^{-1/x}(frac{1}{x^5}+o(x^2)))}}=frac{1}{2}$$
Here is a plot of the function:
answered Jan 19 at 23:38
JimBJimB
55537
$endgroup$
add a comment |
$begingroup$
The Taylor series approach suggested by @Matteo and @WillJagy is the way to go. Here is that approach made a bit more explicit:
- Divide the numerator and denominator by x^6.
- Construct the resulting first order Taylor series for both numerator and denominator.
- Observe what the limits must be.
$$lim_{xto 0^+} , frac{-tan ^{-1}left(2 x^3right)+frac{(x+2) x^5}{x^2+3}+2 x sin ^2(x)}{x exp left(-frac{1}{x}right)-2 x^2+log left(frac{x^2+1}{1-x^2}right)}$$
$$lim_{xto 0^+} ,{ {(frac{1}{3}-frac{2 x}{15}+o(x^2) )}over{(frac{2}{3}+e^{-1/x}(frac{1}{x^5}+o(x^2)))}}=frac{1}{2}$$
Here is a plot of the function:
answered Jan 19 at 23:38
JimBJimB
55537
$endgroup$
The Taylor series approach suggested by @Matteo and @WillJagy is the way to go. Here is that approach made a bit more explicit:
- Divide the numerator and denominator by x^6.
- Construct the resulting first order Taylor series for both numerator and denominator.
- Observe what the limits must be.
$$lim_{xto 0^+} , frac{-tan ^{-1}left(2 x^3right)+frac{(x+2) x^5}{x^2+3}+2 x sin ^2(x)}{x exp left(-frac{1}{x}right)-2 x^2+log left(frac{x^2+1}{1-x^2}right)}$$
$$lim_{xto 0^+} ,{ {(frac{1}{3}-frac{2 x}{15}+o(x^2) )}over{(frac{2}{3}+e^{-1/x}(frac{1}{x^5}+o(x^2)))}}=frac{1}{2}$$
Here is a plot of the function:
answered Jan 19 at 23:38
JimBJimB
55537
edited Jan 19 at 23:46
answered Jan 19 at 23:38
JimBJimB
55537
answered Jan 19 at 23:38
JimBJimB
55537
answered Jan 19 at 23:38
JimBJimB
55537
55537
add a comment |
add a comment |
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$begingroup$
Where did you find this problem
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– Zach
Jan 19 at 19:57
1
$begingroup$
While I know you want to determine the steps to find the limit, that limit (according to Mathematica and a simple plot) is $1/2$.
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– JimB
Jan 19 at 19:58
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@Zach This is question from a previous exam of my course
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– Turan Nəsibli
Jan 19 at 19:58
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You really do need to evaluate each piece up to $x^6.$ There is cancellation among the lower degree terms, including some coefficients that come out $0$ in the first place.
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– Will Jagy
Jan 19 at 20:22
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It looks like $0$: desmos.com/calculator/nrq5rm6ooj
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– clathratus
Jan 19 at 20:44