Mixing
Could someone please explain the steps in the Maclauren series?
$begingroup$
Find the limit of $$lim_{xto 0}{{1-(cos x)({cos2x})^{1/2}(cos3x)^{1/3}}over{x^2}}$$
Steps to solving this problem, from the book:
Step 1: $$lim_{xto 0}{{1-(1-frac{x^2}{2})({1-2x^2})^{1/2}(1-frac{9x^2}{2})^{1/3}}over{x^2}}$$
Step 2: $$lim_{xto 0}{{1-(1-frac{x^2}{2})({1-x^2})(1-frac{3x^2}{2})}over{x^2}}$$
Step 3: $$lim_{xto 0}{{1-(1-frac{3x^2}{2})(1-frac{3x^2}{2})}over{x^2}}$$
Step 4: $$lim_{xto 0}{frac{1-1+3x^2}{x^2}=3 }$$
Could someone please explain to me, what happened in step 2. How did we loose the roots ?
And what happened from step 2 -> step 3 I have no idea what happened there ... :(
limits taylor-expansion limits-without-lhopital
edited Jan 11 at 1:13
Eevee Trainer
5,8421936
asked Jan 11 at 1:12
Nobody1Nobody1
132
$endgroup$
add a comment |
$begingroup$
Find the limit of $$lim_{xto 0}{{1-(cos x)({cos2x})^{1/2}(cos3x)^{1/3}}over{x^2}}$$
Steps to solving this problem, from the book:
Step 1: $$lim_{xto 0}{{1-(1-frac{x^2}{2})({1-2x^2})^{1/2}(1-frac{9x^2}{2})^{1/3}}over{x^2}}$$
Step 2: $$lim_{xto 0}{{1-(1-frac{x^2}{2})({1-x^2})(1-frac{3x^2}{2})}over{x^2}}$$
Step 3: $$lim_{xto 0}{{1-(1-frac{3x^2}{2})(1-frac{3x^2}{2})}over{x^2}}$$
Step 4: $$lim_{xto 0}{frac{1-1+3x^2}{x^2}=3 }$$
Could someone please explain to me, what happened in step 2. How did we loose the roots ?
And what happened from step 2 -> step 3 I have no idea what happened there ... :(
limits taylor-expansion limits-without-lhopital
edited Jan 11 at 1:13
Eevee Trainer
5,8421936
asked Jan 11 at 1:12
Nobody1Nobody1
132
$endgroup$
$begingroup$
Going from step 1 to 2... Your book seems to be using some sort of approximations which ... I don't know if it's legitimate, but it definitely doesn't feel right since I've never quite seen this in the context of limits. It seems even sketchier that they don't openly note that the approximations are being used. In any event, the approximations in question are below. These follow for $x$ that are small in magnitude. $$sqrt{1+x} approx 1 + frac{x}{2}$$ $$sqrt[3]{1+x} approx 1 + frac{x}{3}$$ and generally, for the $n^{th}$ root, $$sqrt[n]{1+x} approx 1 + frac{x}{n}$$
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
Those approximations can be equivalently formulated by $$(1+x)^n approx 1 + nx$$ That said, I'm not sure what happened from step 2 to 3. (As it is I only know about the former approximations because they came up today in my course on modeling.)
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
The title mentions "the Maclauren series", which suggests you are doing a problem connected to a calculus class, but the body of the Question makes no clear connection to what material you might be expected to use to evaluate the limit. Perhaps it is evident to you, but your Readers don't have the advantage of knowing this context.
$endgroup$
– hardmath
Jan 11 at 1:29
add a comment |
$begingroup$
Find the limit of $$lim_{xto 0}{{1-(cos x)({cos2x})^{1/2}(cos3x)^{1/3}}over{x^2}}$$
Steps to solving this problem, from the book:
Step 1: $$lim_{xto 0}{{1-(1-frac{x^2}{2})({1-2x^2})^{1/2}(1-frac{9x^2}{2})^{1/3}}over{x^2}}$$
Step 2: $$lim_{xto 0}{{1-(1-frac{x^2}{2})({1-x^2})(1-frac{3x^2}{2})}over{x^2}}$$
Step 3: $$lim_{xto 0}{{1-(1-frac{3x^2}{2})(1-frac{3x^2}{2})}over{x^2}}$$
Step 4: $$lim_{xto 0}{frac{1-1+3x^2}{x^2}=3 }$$
Could someone please explain to me, what happened in step 2. How did we loose the roots ?
And what happened from step 2 -> step 3 I have no idea what happened there ... :(
limits taylor-expansion limits-without-lhopital
edited Jan 11 at 1:13
Eevee Trainer
5,8421936
asked Jan 11 at 1:12
Nobody1Nobody1
132
$endgroup$
Find the limit of $$lim_{xto 0}{{1-(cos x)({cos2x})^{1/2}(cos3x)^{1/3}}over{x^2}}$$
Steps to solving this problem, from the book:
Step 1: $$lim_{xto 0}{{1-(1-frac{x^2}{2})({1-2x^2})^{1/2}(1-frac{9x^2}{2})^{1/3}}over{x^2}}$$
Step 2: $$lim_{xto 0}{{1-(1-frac{x^2}{2})({1-x^2})(1-frac{3x^2}{2})}over{x^2}}$$
Step 3: $$lim_{xto 0}{{1-(1-frac{3x^2}{2})(1-frac{3x^2}{2})}over{x^2}}$$
Step 4: $$lim_{xto 0}{frac{1-1+3x^2}{x^2}=3 }$$
Could someone please explain to me, what happened in step 2. How did we loose the roots ?
And what happened from step 2 -> step 3 I have no idea what happened there ... :(
limits taylor-expansion limits-without-lhopital
limits taylor-expansion limits-without-lhopital
edited Jan 11 at 1:13
Eevee Trainer
5,8421936
asked Jan 11 at 1:12
Nobody1Nobody1
132
edited Jan 11 at 1:13
Eevee Trainer
5,8421936
asked Jan 11 at 1:12
Nobody1Nobody1
132
edited Jan 11 at 1:13
Eevee Trainer
5,8421936
edited Jan 11 at 1:13
Eevee Trainer
5,8421936
edited Jan 11 at 1:13
Eevee Trainer
5,8421936
5,8421936
asked Jan 11 at 1:12
Nobody1Nobody1
132
asked Jan 11 at 1:12
Nobody1Nobody1
132
asked Jan 11 at 1:12
Nobody1Nobody1
132
132
$begingroup$
Going from step 1 to 2... Your book seems to be using some sort of approximations which ... I don't know if it's legitimate, but it definitely doesn't feel right since I've never quite seen this in the context of limits. It seems even sketchier that they don't openly note that the approximations are being used. In any event, the approximations in question are below. These follow for $x$ that are small in magnitude. $$sqrt{1+x} approx 1 + frac{x}{2}$$ $$sqrt[3]{1+x} approx 1 + frac{x}{3}$$ and generally, for the $n^{th}$ root, $$sqrt[n]{1+x} approx 1 + frac{x}{n}$$
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
Those approximations can be equivalently formulated by $$(1+x)^n approx 1 + nx$$ That said, I'm not sure what happened from step 2 to 3. (As it is I only know about the former approximations because they came up today in my course on modeling.)
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
The title mentions "the Maclauren series", which suggests you are doing a problem connected to a calculus class, but the body of the Question makes no clear connection to what material you might be expected to use to evaluate the limit. Perhaps it is evident to you, but your Readers don't have the advantage of knowing this context.
$endgroup$
– hardmath
Jan 11 at 1:29
add a comment |
$begingroup$
Going from step 1 to 2... Your book seems to be using some sort of approximations which ... I don't know if it's legitimate, but it definitely doesn't feel right since I've never quite seen this in the context of limits. It seems even sketchier that they don't openly note that the approximations are being used. In any event, the approximations in question are below. These follow for $x$ that are small in magnitude. $$sqrt{1+x} approx 1 + frac{x}{2}$$ $$sqrt[3]{1+x} approx 1 + frac{x}{3}$$ and generally, for the $n^{th}$ root, $$sqrt[n]{1+x} approx 1 + frac{x}{n}$$
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
Those approximations can be equivalently formulated by $$(1+x)^n approx 1 + nx$$ That said, I'm not sure what happened from step 2 to 3. (As it is I only know about the former approximations because they came up today in my course on modeling.)
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
The title mentions "the Maclauren series", which suggests you are doing a problem connected to a calculus class, but the body of the Question makes no clear connection to what material you might be expected to use to evaluate the limit. Perhaps it is evident to you, but your Readers don't have the advantage of knowing this context.
$endgroup$
– hardmath
Jan 11 at 1:29
$begingroup$
Going from step 1 to 2... Your book seems to be using some sort of approximations which ... I don't know if it's legitimate, but it definitely doesn't feel right since I've never quite seen this in the context of limits. It seems even sketchier that they don't openly note that the approximations are being used. In any event, the approximations in question are below. These follow for $x$ that are small in magnitude. $$sqrt{1+x} approx 1 + frac{x}{2}$$ $$sqrt[3]{1+x} approx 1 + frac{x}{3}$$ and generally, for the $n^{th}$ root, $$sqrt[n]{1+x} approx 1 + frac{x}{n}$$
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
Going from step 1 to 2... Your book seems to be using some sort of approximations which ... I don't know if it's legitimate, but it definitely doesn't feel right since I've never quite seen this in the context of limits. It seems even sketchier that they don't openly note that the approximations are being used. In any event, the approximations in question are below. These follow for $x$ that are small in magnitude. $$sqrt{1+x} approx 1 + frac{x}{2}$$ $$sqrt[3]{1+x} approx 1 + frac{x}{3}$$ and generally, for the $n^{th}$ root, $$sqrt[n]{1+x} approx 1 + frac{x}{n}$$
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
Those approximations can be equivalently formulated by $$(1+x)^n approx 1 + nx$$ That said, I'm not sure what happened from step 2 to 3. (As it is I only know about the former approximations because they came up today in my course on modeling.)
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
Those approximations can be equivalently formulated by $$(1+x)^n approx 1 + nx$$ That said, I'm not sure what happened from step 2 to 3. (As it is I only know about the former approximations because they came up today in my course on modeling.)
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
The title mentions "the Maclauren series", which suggests you are doing a problem connected to a calculus class, but the body of the Question makes no clear connection to what material you might be expected to use to evaluate the limit. Perhaps it is evident to you, but your Readers don't have the advantage of knowing this context.
$endgroup$
– hardmath
Jan 11 at 1:29
$begingroup$
The title mentions "the Maclauren series", which suggests you are doing a problem connected to a calculus class, but the body of the Question makes no clear connection to what material you might be expected to use to evaluate the limit. Perhaps it is evident to you, but your Readers don't have the advantage of knowing this context.
$endgroup$
– hardmath
Jan 11 at 1:29
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
You should probably get rid of the book. Most calculus texts are like this. The right approach is to use the little / big o notation.
The key here is to use Taylor / Maclaurin expansions as $xto 0$ $$ cos x=1-frac{x^2}{2}+o(x^2)tag{1}$$ and $$(1+x)^n=1+nx+o(x)tag{2}$$ where $o(f(x)) $ represents a function $g(x) $ such that $g(x) /f(x) to 0$.
The starting step (based on expansion $(1)$ above) thus needs to be written like $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-2x^2+o(x^2)right) ^{1/2}left(1-dfrac{9x^2}{2}+o(x^2)right)^{1/3}} {x^2} $$ and the next step (using $(2)$) becomes $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-x^2+o(x^2)right)left(1-dfrac{3x^2}{2}+o(x^2)right)} {x^2} $$ Finally we have via multiplication $$lim_{xto 0}dfrac{1-left(1-3x^2+o(x^2)right)} {x^2}=3 $$
A simpler approach is to use standard limits $$lim_{xto 0}frac{1-cos x} {x^2}=frac{1}{2},,lim_{xto a} frac{x^n-a^n} {x-a} =na^{n-1}$$ which are equivalent to the Taylor expansions given earlier.
The numerator of the expression under limit is of the form $1-abc$ where $a, b, c$ all tend to $1$ as $xto 0$. We can do a split like $$1-abc=1-a+a(1-bc)$$ and thus we have the desired limit as $$lim_{xto 0}frac{1-cos x} {x^2}+lim_{xto 0}cos xcdotfrac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}} {x^2}$$ which is same as $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}}{x^2}$$ Applying the same technique we get $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}}{x^2}+lim_{xto 0}(cos 2x)^{1/2}cdotfrac{1-(cos 3x)^{1/3}}{x^2}$$ The first limit above is evaluated as $$lim_{xto 0}frac{1-(cos 2x)^{1/2}}{1-cos 2x}cdotfrac{1-cos 2x}{(2x)^2}cdot 4=frac{1}{2}cdotfrac{1}{2}cdot 4=1$$ and thus desired limit is equal to $$frac{3}{2}+lim_{xto 0}frac{1-(cos 3x)^{1/3}}{x^2}$$ The last limit is evaluated in similar manner to get $3/2$ and thus the final answer is $3$.
This technique has been used for a more complicated problem here.
answered Jan 11 at 2:11
Paramanand SinghParamanand Singh
50k556163
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add a comment |
$begingroup$
It's the generalized binomial theorem.
In its simplest form,
it states that,
for fixed $a$,
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=0
$.
Informally,
$(1+x)^a approx 1+ax
$
or,
more precisely,
in big-oh notation,
$(1+x)^a = 1+ax+O(x^2)
$.
This follows from
L'Hopital's rule because
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=lim_{x to 0} left(dfrac{a(1+x)^{a-1}-a}{1} right)
=0
$.
answered Jan 11 at 1:21
marty cohenmarty cohen
73.5k549128
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
You should probably get rid of the book. Most calculus texts are like this. The right approach is to use the little / big o notation.
The key here is to use Taylor / Maclaurin expansions as $xto 0$ $$ cos x=1-frac{x^2}{2}+o(x^2)tag{1}$$ and $$(1+x)^n=1+nx+o(x)tag{2}$$ where $o(f(x)) $ represents a function $g(x) $ such that $g(x) /f(x) to 0$.
The starting step (based on expansion $(1)$ above) thus needs to be written like $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-2x^2+o(x^2)right) ^{1/2}left(1-dfrac{9x^2}{2}+o(x^2)right)^{1/3}} {x^2} $$ and the next step (using $(2)$) becomes $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-x^2+o(x^2)right)left(1-dfrac{3x^2}{2}+o(x^2)right)} {x^2} $$ Finally we have via multiplication $$lim_{xto 0}dfrac{1-left(1-3x^2+o(x^2)right)} {x^2}=3 $$
A simpler approach is to use standard limits $$lim_{xto 0}frac{1-cos x} {x^2}=frac{1}{2},,lim_{xto a} frac{x^n-a^n} {x-a} =na^{n-1}$$ which are equivalent to the Taylor expansions given earlier.
The numerator of the expression under limit is of the form $1-abc$ where $a, b, c$ all tend to $1$ as $xto 0$. We can do a split like $$1-abc=1-a+a(1-bc)$$ and thus we have the desired limit as $$lim_{xto 0}frac{1-cos x} {x^2}+lim_{xto 0}cos xcdotfrac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}} {x^2}$$ which is same as $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}}{x^2}$$ Applying the same technique we get $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}}{x^2}+lim_{xto 0}(cos 2x)^{1/2}cdotfrac{1-(cos 3x)^{1/3}}{x^2}$$ The first limit above is evaluated as $$lim_{xto 0}frac{1-(cos 2x)^{1/2}}{1-cos 2x}cdotfrac{1-cos 2x}{(2x)^2}cdot 4=frac{1}{2}cdotfrac{1}{2}cdot 4=1$$ and thus desired limit is equal to $$frac{3}{2}+lim_{xto 0}frac{1-(cos 3x)^{1/3}}{x^2}$$ The last limit is evaluated in similar manner to get $3/2$ and thus the final answer is $3$.
This technique has been used for a more complicated problem here.
answered Jan 11 at 2:11
Paramanand SinghParamanand Singh
50k556163
$endgroup$
add a comment |
$begingroup$
You should probably get rid of the book. Most calculus texts are like this. The right approach is to use the little / big o notation.
The key here is to use Taylor / Maclaurin expansions as $xto 0$ $$ cos x=1-frac{x^2}{2}+o(x^2)tag{1}$$ and $$(1+x)^n=1+nx+o(x)tag{2}$$ where $o(f(x)) $ represents a function $g(x) $ such that $g(x) /f(x) to 0$.
The starting step (based on expansion $(1)$ above) thus needs to be written like $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-2x^2+o(x^2)right) ^{1/2}left(1-dfrac{9x^2}{2}+o(x^2)right)^{1/3}} {x^2} $$ and the next step (using $(2)$) becomes $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-x^2+o(x^2)right)left(1-dfrac{3x^2}{2}+o(x^2)right)} {x^2} $$ Finally we have via multiplication $$lim_{xto 0}dfrac{1-left(1-3x^2+o(x^2)right)} {x^2}=3 $$
A simpler approach is to use standard limits $$lim_{xto 0}frac{1-cos x} {x^2}=frac{1}{2},,lim_{xto a} frac{x^n-a^n} {x-a} =na^{n-1}$$ which are equivalent to the Taylor expansions given earlier.
The numerator of the expression under limit is of the form $1-abc$ where $a, b, c$ all tend to $1$ as $xto 0$. We can do a split like $$1-abc=1-a+a(1-bc)$$ and thus we have the desired limit as $$lim_{xto 0}frac{1-cos x} {x^2}+lim_{xto 0}cos xcdotfrac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}} {x^2}$$ which is same as $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}}{x^2}$$ Applying the same technique we get $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}}{x^2}+lim_{xto 0}(cos 2x)^{1/2}cdotfrac{1-(cos 3x)^{1/3}}{x^2}$$ The first limit above is evaluated as $$lim_{xto 0}frac{1-(cos 2x)^{1/2}}{1-cos 2x}cdotfrac{1-cos 2x}{(2x)^2}cdot 4=frac{1}{2}cdotfrac{1}{2}cdot 4=1$$ and thus desired limit is equal to $$frac{3}{2}+lim_{xto 0}frac{1-(cos 3x)^{1/3}}{x^2}$$ The last limit is evaluated in similar manner to get $3/2$ and thus the final answer is $3$.
This technique has been used for a more complicated problem here.
answered Jan 11 at 2:11
Paramanand SinghParamanand Singh
50k556163
$endgroup$
add a comment |
$begingroup$
You should probably get rid of the book. Most calculus texts are like this. The right approach is to use the little / big o notation.
The key here is to use Taylor / Maclaurin expansions as $xto 0$ $$ cos x=1-frac{x^2}{2}+o(x^2)tag{1}$$ and $$(1+x)^n=1+nx+o(x)tag{2}$$ where $o(f(x)) $ represents a function $g(x) $ such that $g(x) /f(x) to 0$.
The starting step (based on expansion $(1)$ above) thus needs to be written like $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-2x^2+o(x^2)right) ^{1/2}left(1-dfrac{9x^2}{2}+o(x^2)right)^{1/3}} {x^2} $$ and the next step (using $(2)$) becomes $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-x^2+o(x^2)right)left(1-dfrac{3x^2}{2}+o(x^2)right)} {x^2} $$ Finally we have via multiplication $$lim_{xto 0}dfrac{1-left(1-3x^2+o(x^2)right)} {x^2}=3 $$
A simpler approach is to use standard limits $$lim_{xto 0}frac{1-cos x} {x^2}=frac{1}{2},,lim_{xto a} frac{x^n-a^n} {x-a} =na^{n-1}$$ which are equivalent to the Taylor expansions given earlier.
The numerator of the expression under limit is of the form $1-abc$ where $a, b, c$ all tend to $1$ as $xto 0$. We can do a split like $$1-abc=1-a+a(1-bc)$$ and thus we have the desired limit as $$lim_{xto 0}frac{1-cos x} {x^2}+lim_{xto 0}cos xcdotfrac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}} {x^2}$$ which is same as $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}}{x^2}$$ Applying the same technique we get $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}}{x^2}+lim_{xto 0}(cos 2x)^{1/2}cdotfrac{1-(cos 3x)^{1/3}}{x^2}$$ The first limit above is evaluated as $$lim_{xto 0}frac{1-(cos 2x)^{1/2}}{1-cos 2x}cdotfrac{1-cos 2x}{(2x)^2}cdot 4=frac{1}{2}cdotfrac{1}{2}cdot 4=1$$ and thus desired limit is equal to $$frac{3}{2}+lim_{xto 0}frac{1-(cos 3x)^{1/3}}{x^2}$$ The last limit is evaluated in similar manner to get $3/2$ and thus the final answer is $3$.
This technique has been used for a more complicated problem here.
answered Jan 11 at 2:11
Paramanand SinghParamanand Singh
50k556163
$endgroup$
You should probably get rid of the book. Most calculus texts are like this. The right approach is to use the little / big o notation.
The key here is to use Taylor / Maclaurin expansions as $xto 0$ $$ cos x=1-frac{x^2}{2}+o(x^2)tag{1}$$ and $$(1+x)^n=1+nx+o(x)tag{2}$$ where $o(f(x)) $ represents a function $g(x) $ such that $g(x) /f(x) to 0$.
The starting step (based on expansion $(1)$ above) thus needs to be written like $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-2x^2+o(x^2)right) ^{1/2}left(1-dfrac{9x^2}{2}+o(x^2)right)^{1/3}} {x^2} $$ and the next step (using $(2)$) becomes $$lim_{xto 0}dfrac{1-left(1-dfrac {x^2}{2}+o(x^2)right)left(1-x^2+o(x^2)right)left(1-dfrac{3x^2}{2}+o(x^2)right)} {x^2} $$ Finally we have via multiplication $$lim_{xto 0}dfrac{1-left(1-3x^2+o(x^2)right)} {x^2}=3 $$
A simpler approach is to use standard limits $$lim_{xto 0}frac{1-cos x} {x^2}=frac{1}{2},,lim_{xto a} frac{x^n-a^n} {x-a} =na^{n-1}$$ which are equivalent to the Taylor expansions given earlier.
The numerator of the expression under limit is of the form $1-abc$ where $a, b, c$ all tend to $1$ as $xto 0$. We can do a split like $$1-abc=1-a+a(1-bc)$$ and thus we have the desired limit as $$lim_{xto 0}frac{1-cos x} {x^2}+lim_{xto 0}cos xcdotfrac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}} {x^2}$$ which is same as $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}(cos 3x)^{1/3}}{x^2}$$ Applying the same technique we get $$frac{1}{2}+lim_{xto 0}frac{1-(cos 2x)^{1/2}}{x^2}+lim_{xto 0}(cos 2x)^{1/2}cdotfrac{1-(cos 3x)^{1/3}}{x^2}$$ The first limit above is evaluated as $$lim_{xto 0}frac{1-(cos 2x)^{1/2}}{1-cos 2x}cdotfrac{1-cos 2x}{(2x)^2}cdot 4=frac{1}{2}cdotfrac{1}{2}cdot 4=1$$ and thus desired limit is equal to $$frac{3}{2}+lim_{xto 0}frac{1-(cos 3x)^{1/3}}{x^2}$$ The last limit is evaluated in similar manner to get $3/2$ and thus the final answer is $3$.
This technique has been used for a more complicated problem here.
answered Jan 11 at 2:11
Paramanand SinghParamanand Singh
50k556163
edited Jan 11 at 2:32
answered Jan 11 at 2:11
Paramanand SinghParamanand Singh
50k556163
answered Jan 11 at 2:11
Paramanand SinghParamanand Singh
50k556163
answered Jan 11 at 2:11
Paramanand SinghParamanand Singh
50k556163
50k556163
add a comment |
add a comment |
$begingroup$
It's the generalized binomial theorem.
In its simplest form,
it states that,
for fixed $a$,
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=0
$.
Informally,
$(1+x)^a approx 1+ax
$
or,
more precisely,
in big-oh notation,
$(1+x)^a = 1+ax+O(x^2)
$.
This follows from
L'Hopital's rule because
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=lim_{x to 0} left(dfrac{a(1+x)^{a-1}-a}{1} right)
=0
$.
answered Jan 11 at 1:21
marty cohenmarty cohen
73.5k549128
$endgroup$
add a comment |
$begingroup$
It's the generalized binomial theorem.
In its simplest form,
it states that,
for fixed $a$,
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=0
$.
Informally,
$(1+x)^a approx 1+ax
$
or,
more precisely,
in big-oh notation,
$(1+x)^a = 1+ax+O(x^2)
$.
This follows from
L'Hopital's rule because
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=lim_{x to 0} left(dfrac{a(1+x)^{a-1}-a}{1} right)
=0
$.
answered Jan 11 at 1:21
marty cohenmarty cohen
73.5k549128
$endgroup$
add a comment |
$begingroup$
It's the generalized binomial theorem.
In its simplest form,
it states that,
for fixed $a$,
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=0
$.
Informally,
$(1+x)^a approx 1+ax
$
or,
more precisely,
in big-oh notation,
$(1+x)^a = 1+ax+O(x^2)
$.
This follows from
L'Hopital's rule because
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=lim_{x to 0} left(dfrac{a(1+x)^{a-1}-a}{1} right)
=0
$.
answered Jan 11 at 1:21
marty cohenmarty cohen
73.5k549128
$endgroup$
It's the generalized binomial theorem.
In its simplest form,
it states that,
for fixed $a$,
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=0
$.
Informally,
$(1+x)^a approx 1+ax
$
or,
more precisely,
in big-oh notation,
$(1+x)^a = 1+ax+O(x^2)
$.
This follows from
L'Hopital's rule because
$lim_{x to 0} left(dfrac{(1+x)^a-(1+ax)}{x} right)
=lim_{x to 0} left(dfrac{a(1+x)^{a-1}-a}{1} right)
=0
$.
answered Jan 11 at 1:21
marty cohenmarty cohen
73.5k549128
answered Jan 11 at 1:21
marty cohenmarty cohen
73.5k549128
answered Jan 11 at 1:21
marty cohenmarty cohen
73.5k549128
answered Jan 11 at 1:21
marty cohenmarty cohen
73.5k549128
73.5k549128
add a comment |
add a comment |
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$begingroup$
Going from step 1 to 2... Your book seems to be using some sort of approximations which ... I don't know if it's legitimate, but it definitely doesn't feel right since I've never quite seen this in the context of limits. It seems even sketchier that they don't openly note that the approximations are being used. In any event, the approximations in question are below. These follow for $x$ that are small in magnitude. $$sqrt{1+x} approx 1 + frac{x}{2}$$ $$sqrt[3]{1+x} approx 1 + frac{x}{3}$$ and generally, for the $n^{th}$ root, $$sqrt[n]{1+x} approx 1 + frac{x}{n}$$
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
Those approximations can be equivalently formulated by $$(1+x)^n approx 1 + nx$$ That said, I'm not sure what happened from step 2 to 3. (As it is I only know about the former approximations because they came up today in my course on modeling.)
$endgroup$
– Eevee Trainer
Jan 11 at 1:19
$begingroup$
The title mentions "the Maclauren series", which suggests you are doing a problem connected to a calculus class, but the body of the Question makes no clear connection to what material you might be expected to use to evaluate the limit. Perhaps it is evident to you, but your Readers don't have the advantage of knowing this context.
$endgroup$
– hardmath
Jan 11 at 1:29