L'Hospital Rule With Exponents - $\tan x/x$

given the following limit:

$$\lim _ {x \to 0 } \left (\frac{\tan x } {x} \right ) ^{1/x^2}\;,$$

is there any simple way to calculate it ?

I have tried writing it as $e^ {\ln (\dots)}$ , but it doesn't give me anything [ and I did l'Hospital on the limit I received... ]

can someone help me with this?

thanks a lot !

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Rewriting as $e^{\ln ...}$ is a good idea. Maybe you should apply l'Hospital it twice. –  Fabian Dec 31 '12 at 16:34
applying it twice also doesn't help ... So my guess is that there is some other way to do it . –  theMissingIngredient Dec 31 '12 at 16:35
There is some other way to do it. Do you know Taylor expansion? –  Fabian Dec 31 '12 at 16:36
Yes, but I am pretty sure we need to do it with l'Hospital Is there any trick to solve it with l'Hospital ? Thanks ! –  theMissingIngredient Dec 31 '12 at 16:45

If you know Taylor expansion, you know that $$\tan x = x + \frac{x^3}{3}+ \mathcal{O}(x^5)$$ where the big-Oh denotes a term which scales like $x^5$ for $x\to 0$. Thus, $$\frac{\tan x}{x} = 1 + \frac{x^2}{3} + \mathcal{O}(x^4).$$ The expansion of the logarithm around $1$ reads $$\ln (1+y) = y + \mathcal{O}(y^2).$$ Letting $1+y=\tan x/x =1+ x^2/3 + \mathcal{O}(x^4)$, we obtain $$\ln \left(\frac{\tan x}x \right) = \frac{x^2}3 + \mathcal{O}(x^4).$$ Now, $$\frac1 {x^2} \ln \left(\frac{\tan x}x \right) = \frac13 + \mathcal{O}(x^2).$$ And thus $$\lim_{x\to 0} \frac1 {x^2} \ln \left(\frac{\tan x}x \right) = \frac13.$$

With that you can easily show that $$\lim_{x\to 0} \left(\frac{\tan x}x\right) ^{x^{-2}} = \sqrt[3]{e}.$$

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thanks a lot !!!!!!!!!! –  theMissingIngredient Dec 31 '12 at 17:10
just one thing: in your 9th line, you wrote $ln (1+\frac{x^2 }{3} +O(x^4)$ = $\frac{x^2 }{3} +O(x^4)$ . Shouldn't you also change something in the big-o ? thanks!@ –  theMissingIngredient Dec 31 '12 at 17:20
i.e.- your $y$ is $y = \frac{x^2}{3} +O(x^4 )$ , and you are looking at $ln(1+y)$ –  theMissingIngredient Dec 31 '12 at 17:20
The next term in the expansion of $\ln (1+y)= x^2/3+ ...$ comes from the first order term of $O(x^4)$ and the second order term of $x^2/3$ both of which are of order $O(x^4)$. (hope this is understandable) –  Fabian Dec 31 '12 at 17:25
Soryy, but I couldn't understand ... Shouldn't you have $O(x^8)$ after substituting? –  theMissingIngredient Dec 31 '12 at 18:37

$$\lim _ {x \to 0 } \left (\frac{\tan x } {x} \right ) ^{1/x^2}=\lim_{x\to 0}\left(1+\left({\frac{\tan x-x}{x}}\right)\right)^{{x\over\tan x-x}{\tan x-x\over x^3}}=e^{\lim_{x\to 0}\frac{\tan x-x}{x^3}}$$

$$\lim_{x\to 0}\frac{\tan x-x}{x^3}=\lim_{x\to 0}\frac{{1\over \cos^2x}-1}{3x^2}={1\over 3}\lim_{x\to 0}\left({\tan x\over x}\right)^2={1\over 3}$$ $$\lim _ {x \to 0 } \left (\frac{\tan x } {x} \right ) ^{1/x^2}=e^{1/3}$$

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How good do you see my approach, Adi?:) –  B. S. Jan 1 '13 at 7:35

Use this fact as well:

If $\lim\limits_{x\to{+\infty}} f(x)^{g(x)}$ be as $1^{+\infty}$, which is an indeterminate form, then we have this fact that: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim\limits_{x\to +\infty}\big(f(x)-1\big)g(x)}$$

Here we have $$\lim_{x\to+\infty}\exp\left(\frac{\tan(x)-x}{x^3}\right)=\exp(1/3)$$

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Helpful to know! +1 –  amWhy Feb 26 '13 at 0:54
@amWhy: Thanks ;-) –  B. S. Feb 26 '13 at 2:58
I really like how you contribute, giving diverse ways of looking at problems, and doing so in a helpful, nudging, and instructive way. Students so often fall into the "trap" of thinking there is only one correct answer, or one way of doing things, etc...so you help bring "creativity" to math! –  amWhy Feb 26 '13 at 3:01
Thanks for saying so. It is very kind of you @amWhy. :-) –  B. S. Feb 26 '13 at 3:03

It is possible to do it with l'Hospital's rule. It takes 4 applications, but it does work! Do the exponential transformation, and continue simplifying with l'Hospital's rule and limits until you get:

$$\exp\left(-\frac{\left(2 \left(\lim_{x\rightarrow0} \cos\left(2 x\right)\right)\right)}{\left(\lim_{x\rightarrow0} \left(\left(4 x^2-6\right) \cos\left(2 x\right)+12 x \sin\left(2 x\right)\right)\right)}\right)$$

Use a bunch of limit rules (quotient, continuity, sum, product, polynomial, in that order), in order to get: $$\exp\left(-\frac{\left(2 \cos\left(\lim_{x\rightarrow0} 2 x\right)\right)}{\left(12 \left(\lim_{x\rightarrow0} x\right) \left(\lim_{x\rightarrow0} \sin\left(2 x\right)\right)-6 \cos\left(\lim_{x\rightarrow0} 2 x\right)\right)}\right)$$

Just evaluate all the limits now: $$\exp\left(\frac{1}{3}\right)$$

Tedious, but certainly doable! I would recommend Fabian's solution instead, 4 applications of Hospital's, although possible, is something you want to avoid if possible.

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Let $$y=\left (\frac{\tan x } {x} \right ) ^{1/x^2}$$

So, $$\ln y=\frac{\ln \tan x -\ln x}{x^2}$$

Then $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln \tan x -\ln x}{x^2}\left(\frac 00\right)$$ as $\lim_{x\to 0}\frac {\tan x}x=1$

Applying L'Hospital's Rule: , $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\frac2{\sin2x}-\frac1x}{2x}=\lim_{x\to 0}\frac{2x-\sin2x}{2x^2\sin2x}\left(\frac 00\right)$$

$$=\lim_{x\to 0}\frac{2-2\cos2x}{4x\sin2x+4x^2\cos 2x}\left(\frac 00\right)$$ (applying L'Hospital's Rule)

$$=\lim_{x\to 0}\frac{4\sin2x}{4\sin2x+8x\cos2x+8x\cos 2x-8x^2\sin2x}\left(\frac 00\right)$$ (applying L'Hospital's Rule)

$$=\lim_{x\to 0}\frac{8\cos2x}{8\cos2x+2(8\cos2x-16x\sin2x)-16x\sin2x-16x^2\cos2x}\left(\frac 00\right)$$ (applying L'Hospital's Rule)

$$=\frac8{8+2\cdot8}=\frac13$$

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wow ! great ! thanks a lot !!! –  theMissingIngredient Dec 31 '12 at 18:40
@theMissingIngredient, my pleasure. I think the problem is best solved by Fabian's way if 'L'Hospital's Rule' is not mentioned. –  lab bhattacharjee Dec 31 '12 at 18:43

One might rewrite @Fabian's answer without big-O thus:

Since $\tan x = x + {x^3\over3} + {2x^5\over15}+\cdots$ and $\log(1+x) = x - {x^2\over2} + \cdots$, we have $$\left({\tan x\over x}\right)^{1/x^2} =e^{{1\over x^2}\log\left({\tan x \over x}\right)} =e^{{1\over x^2}\log\left(1+{x^2\over3}+{2x^4\over15}+\cdots\right)} =e^{{1\over x^2}\left(\left({x^2\over3}+{2x^4\over15}+\cdots\right)-{1\over2}\left({x^2\over3}+{2x^4\over15}+\cdots\right)^2+\cdots\right)} =e^{{1\over3}+{7x^2\over90}+\cdots}$$ which approaches $\root 3 \of e$.

However in a calculus course based on standard US textbooks, the problem looks like a standard, if involved, L'Hopital's Rule problem, which can be handled as @lab bhattacharjee, with one simplification thus (all limits as $x\rightarrow0$): $$\log\lim \left({\tan x\over x}\right)^{1/x^2} = \lim {\log\left({\tan x\over x}\right)\over{x^2}} \buildrel L'H \over = \lim {2x -\sin 2x \over 2x^2 \sin2x} = \lim {2x -\sin 2x \over 4x^3} \lim {2x \over \sin2x} \buildrel L'H \over = \lim {2\over3}{1 -\cos 2x \over (2x)^2} = {1 \over3}\,,$$ the last limit either being known to be $1/2$ or done by L'Hopital's rule twice.

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Great !! Thanks a lot !!! –  theMissingIngredient Jan 1 '13 at 21:02