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Calculate the following limit: $$\lim_{x \rightarrow 0} \left( \frac{\tan x}{x} \right) ^ \frac{1}{\sin^2 x}$$

I know the result must be $\sqrt[3]{e}$ but I don't know how to get it. I've tried rewriting the limit as follows:

$$\lim_{x \rightarrow 0} e ^ {\ln {\left( \frac{\tan x}{x} \right) ^ \frac{1}{\sin^2 x}}} = \lim_{x \rightarrow 0} e ^ {\frac{1}{\sin^2 x} \ln {\left( \frac{\tan x}{x} \right)}}$$

From this point, I applied l'Hospital's rule but got $1$ instead of $\sqrt[3]{e}$.

Thank you!

George R.
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4 Answers4

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$$\lim_{x\to0}\frac{\log\frac{\tan x}x}{\sin^2x}\stackrel{l'H}=\lim_{x\to0}\frac{\frac x{\tan x}\frac{x\sec^2x-\tan x}{x^2}}{2\sin x\cos x}=\lim_{x\to0}\frac {\frac1{\sin x\cos x}-\frac1x}{2\sin x\cos x}=$$

$$=\lim_{x\to0}\frac{x-\sin x\cos x}{\underbrace{2x\sin^2x\cos^2x}_{=\frac x2\sin^22x}}\stackrel{l'H}=\lim_{x\to0}\frac{\overbrace{1-\cos^2 x+\sin^2x}^{2\sin^2x}}{\frac12\sin^22x+\underbrace{x\sin2x\cos2x}_{=x\sin4x}}\stackrel{l'H}=\lim_{x\to0}\frac{2\sin2x}{2\sin4x+4x\cos4x}=$$

$$\stackrel{l'H}=\lim_{x\to0}\frac{4\cos2x}{12\cos4x-16x\sin4x}=\frac4{12}=\frac13$$

and the limit is $\;\;e^{1/3}\;$

DonAntonio
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Lemma: Suppose $\lim\limits_{x\to0}xy(x)=a$, then $$ \lim_{x\to0}(1+x)^y=e^a\tag{1} $$ Proof: For any $\epsilon\gt0$, there is a $\delta\gt0$ so that if $|x|\le\delta$, then $$ a-\epsilon\le xy\le a+\epsilon\tag{2} $$ Then, for $|x|\le\delta$, $$ (1+x)^{\frac{a-\epsilon}x}\le(1+x)^y\le(1+x)^{\frac{a+\epsilon}x}\tag{3} $$ and taking the limit of $(3)$ as $x\to0$, we get $$ e^{a-\epsilon}\le\lim_{x\to0}(1+x)^y\le e^{a+\epsilon}\tag{4} $$ Since $(4)$ is true for any $\epsilon\gt0$, we have $(1)$.

QED

As shown in this answer, $$ \lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)}=-\frac12\tag{5} $$ Applying $\frac1{1-x}$, which is continuous at $x=-\frac12$, to $(5)$ yields $$ \lim_{x\to0}\frac{\tan(x)-x}{\tan(x)-\sin(x)}=\frac23\tag{6} $$ Since $\frac{\tan(x)-\sin(x)}{\sin^3(x)}=\frac1{\cos(x)(\cos(x)+1)}$, we get $$ \begin{align} \lim_{x\to0}\frac{\tan(x)-x}{x\sin^2(x)} &=\left(\lim_{x\to0}\frac{\tan(x)-x}{\tan(x)-\sin(x)}\right)\left(\lim_{x\to0}\frac{\tan(x)-\sin(x)}{\sin^3(x)}\right)\left(\lim_{x\to0}\frac{\sin(x)}x\right)\\ &=\frac23\cdot\frac12\cdot1\\[3pt] &=\frac13\tag{7} \end{align} $$

Therefore, applying $(1)$ and $(7)$, we get $$ \begin{align} \lim_{x\to0}\left(\frac{\tan(x)}x\right)^{1/\sin^2(x)} &=\lim_{x\to0}\left(1+\frac{\tan(x)-x}x\right)^{1/\sin^2(x)}\\[6pt] &=e^{1/3}\tag{8} \end{align} $$

Community
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robjohn
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  • It's normally not advised to solve indeterminates of the form $1^{\infty }$ by taking limits of each part. – PenasRaul Nov 02 '16 at 12:20
  • @Rob I agree with the past comment (though the downvote, as usual, seems rushed up...). In one before the last line you took the limit of the red part separatedly and inside the parentheses. -- and also in the exponent!-- , something that would cost dearly to my students if they'd ever dare to do such a thing...without justification, of course. – DonAntonio Nov 02 '16 at 12:25
  • I have added an explanation and a bit of clarification to $(2)$. – robjohn Nov 02 '16 at 12:45
  • @R Very nice work explaining that step, yet I think that at the end of your post, it is not that "we have (2)", as we go back to a potentially problematic thing after "disappearing" the $;\frac{tan x-x}{x^3};$ expression within the parentheses. I think that after reaching point (1) you can use then the last part of your answer, with all the $;x'$ s and $;\epsilon,$'s and the inequalities in the whole expression, without taking partially the limit when $;x\to0;$ , and then pass to the limit in both extremes, use squeeze theorem and etc. – DonAntonio Nov 02 '16 at 12:46
  • @DonAntonio: I have split the argument into two parts since the original presentation seems to have been confusing. There is a lack of notation for between-ness. – robjohn Nov 02 '16 at 13:00
  • I have made a major update to the presentation, but not the spirit of the answer. I can understand the downvote from before I added an explanation, but with the new presentation, does the downvote still apply? If so, would you like to explain? – robjohn Nov 02 '16 at 18:49
  • @PenasRaul: I have added the Lemma which I was implicitly using. I have also added a proof without L'Hôpital of $(7)$. – robjohn Nov 04 '16 at 17:59
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You have an indeterminate form of kind $1^{\infty}.$ You can solve this type of problems via:

$$\lim_{x\to a} f(x)^{g(x)}=e^{\lim_{x\to a} (f(x)-1)g(x)}.$$

Edit

If $\lim_{x\to a}f(x)=1$ then it is $$\lim_{x\to a} f(x)^{g(x)}=\lim_{x\to a} (1+f(x)-1)^{\dfrac{f(x)-1}{f(x-1)}g(x)}=\lim_{x\to a} \left( (1+f(x)-1)^{\dfrac{1}{f(x)-1}}\right)^{f(x-1)g(x)}.$$

Now, it is $$\lim_{x\to a} \left( (1+f(x)-1)^{\dfrac{1}{f(x)-1}}\right)=e.$$

So, if $\lim_{x\to a} (f(x)-1)g(x)$ exists we have $$\lim_{x\to a} f(x)^{g(x)}=e^{\lim_{x\to a} (f(x)-1)g(x)}.$$

End of the edit

In this case,

$$\begin{align}\lim_{x\to 0}\left(\dfrac{\tan x}{x}-1\right)\dfrac{1}{\sin^2 x} & \\ &= \lim_{x\to 0}\dfrac{\tan x-x}{x\sin^2 x} \\ &=\lim_{x\to 0}\dfrac{\dfrac{1}{\cos^2 x}-1}{\sin^2 x+2x\sin x\cos x} \\& =\lim_{x\to 0}\dfrac{\sin^2 x}{\cos^2 x(\sin^2 x+2x\sin x\cos x)}\\&=\lim_{x\to 0}\dfrac{\sin x}{\cos^2 x(\sin x+2x\cos x) }\\& =\lim_{x\to 0}\dfrac{\cos x}{-2\cos x\sin x(\sin x+2x\cos x)+\cos^2x(3\cos x-2x\sin x) }\\&=\dfrac 13.\end{align}$$

mfl
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    Does this rule always apply? Am I interpreting it correctly by saying that at you are only keeping the first term in the Taylor expansion of $$\log(f(x))=\log(1+f(x)-1)?$$ – b00n heT Nov 02 '16 at 12:18
  • I also think the step b00n is talking about requires justification...at least some link to it. – DonAntonio Nov 02 '16 at 12:28
  • @DonAntonio I have editted the answer to justify my claim. – mfl Nov 02 '16 at 13:18
  • @b00nheT I have editted the answer to justify my claim. – mfl Nov 02 '16 at 13:18
  • @mfl Thanks. In your edit perhaps it could be added that it must be $;f(x)\rightarrow[x\to a]{}1;$ . Anyway, it doesn't seem to be this way is a big time or difficulty saver when compared with other methods. – DonAntonio Nov 02 '16 at 13:29
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To evaluate $$ \lim_{x\rightarrow 0}\left(\frac{\tan(x)}{x}\right)^{\frac{1}{\sin^2(x)}} $$ we first observe that $\lim_{x\rightarrow 0}\frac{\tan(x)}{x}=\lim_{x\rightarrow 0}\frac{1}{\cos(x)}\frac{\sin(x)}{x}$. Both of these factors approach $1$ as $x$ approaches $0$. On the other hand, since $\sin(x)$ approaches $0$ as $x$ approaches $0$, this limit is of the form $1^\infty$. We now consider exponentiation:

$$ \lim_{x\rightarrow 0}\left(\frac{\tan(x)}{x}\right)^{\frac{1}{\sin^2(x)}} =\operatorname{exp}\left(\lim_{x\rightarrow 0}\frac{1}{\sin^2(x)}\ln\left(\frac{\tan(x)}{x}\right)\right). $$ We now investigate the limit: $$ \lim_{x\rightarrow 0}\frac{1}{\sin^2(x)}\ln\left(\frac{\tan(x)}{x}\right). $$ Since we have seen that $\frac{\tan(x)}{x}$ approaches $1$, the logarithm approaches $0$, so this is of indeterminate form $\frac{0}{0}$ and l'Hopital's rule applies. Therefore, \begin{align*} \lim_{x\rightarrow 0}\frac{1}{\sin^2(x)}\ln\left(\frac{\tan(x)}{x}\right)&= \lim_{x\rightarrow 0}\frac{1}{2\sin(x)\cos(x)}\frac{1}{\frac{\tan(x)}{x}}\frac{x\cos^2(x)-\sin(x)(\cos(x)-x\sin(x))}{x^2}\\ &=\lim_{x\rightarrow 0}\frac{x-\sin(x)\cos(x)}{2x\sin^2(x)} \end{align*} This is, again, an indeterminate form of $\frac{0}{0}$, so we can apply l'Hopital's rule again to get \begin{align*} \lim_{x\rightarrow 0}\frac{x-\sin(x)\cos(x)}{2x\sin^2(x)}&=\lim_{x\rightarrow 0}\frac{1-\cos^2(x)+\sin^2(x)}{2\sin^2(x)+2x\sin(x)\cos(x)}\\ &=\lim_{x\rightarrow 0}\frac{2\sin^2(x)}{2\sin^2(x)+4x\sin(x)\cos(x)}\\ &=\lim_{x\rightarrow 0}\frac{\sin(x)}{\sin(x)+2x\cos(x)} \end{align*} Once again, this is of an indeterminate form $\frac{0}{0}$. So, we apply l'Hopital's rule one last time to get \begin{align*} \lim_{x\rightarrow 0}\frac{\sin(x)}{\sin(x)+2x\cos(x)}&=\lim_{x\rightarrow 0}\frac{\cos(x)}{\cos(x)+2\cos(x)-2x\sin(x)}=\frac{1}{3}. \end{align*} Therefore, the original limit is $e^{1/3}$.

Michael Burr
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