Calculate exponential limit involving trigonometric functions

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!

• L'Hopital's rule can fail, but it shouldn't give the wrong answer. Can you provide your work? – Michael Burr Nov 2 '16 at 12:05
• – Workaholic Nov 2 '16 at 15:18

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

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

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

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

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 $e^{1/3}$.

• This is exactly what I did almost 40 minutes before your post. Perhaps I missed something, but does this add something that other answers didn't have? – DonAntonio Nov 2 '16 at 13:43
• @DonAntonio This doesn't require double angle formulas (which are often forgotten). – Michael Burr Nov 2 '16 at 14:11
• @DonAntonio Other than avoiding the double angle formulas, the approaches are the same. – Michael Burr Nov 2 '16 at 14:17

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}

• 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 2 '16 at 12:18
• I also think the step b00n is talking about requires justification...at least some link to it. – DonAntonio Nov 2 '16 at 12:28
• @DonAntonio I have editted the answer to justify my claim. – mfl Nov 2 '16 at 13:18
• @b00nheT I have editted the answer to justify my claim. – mfl Nov 2 '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 2 '16 at 13:29

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}

• It's normally not advised to solve indeterminates of the form $1^{\infty }$ by taking limits of each part. – PenasRaul Nov 2 '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 2 '16 at 12:25
• I have added an explanation and a bit of clarification to $(2)$. – robjohn Nov 2 '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 2 '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 2 '16 at 13:00