Question

29" id="TexFormula1" title="\lim_{n \to \0}(x/(tan⁡(x))^(cot⁡(x)^2 )" alt="\lim_{n \to \0}(x/(tan⁡(x))^(cot⁡(x)^2 )" align="absmiddle" class="latex-formula">

Answer 1

It looks like the limit you want to compute is

$\displaystyle L = \lim_{x\to0}\left(\frac x{\tan(x)}\right)^{\cot^2(x)}$

Rewrite the limand with an exponential and logarithm:

$\left(\dfrac{x}{\tan(x)}\right)^{\cot^2(x)} = \exp\left(\cot^2(x) \ln\left(\dfrac{x}{\tan(x)}\right)\right) = \exp\left(\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)}\right)$

Now, since the exponential function is continuous at 0, we can write

$\displaystyle L = \lim_{x\to0} \exp\left(\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)}\right) = \exp\left(\lim_{x\to0}\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)}\right)$

Let M denote the remaining limit.

We have $\dfrac x{\tan(x)}\to1$ as $x\to0$, so $\ln\left(\dfrac x{\tan(x)}\right)\to0$ and $\tan^2(x)\to0$. Apply L'Hopital's rule:

$\displaystyle M = \lim_{x\to0}\dfrac{\ln\left(\dfrac{x}{\tan(x)}\right)}{\tan^2(x)} \\\\ M = \lim_{x\to0}\dfrac{\dfrac{\tan(x)-x\sec^2(x)}{\tan^2(x)}\times\dfrac{\tan(x)}{x}}{2\tan(x)\sec^2(x)}$

Simplify and rewrite this in terms of sin and cos :

$\displaystyle M = \lim_{x\to0} \dfrac{\dfrac{\tan(x)-x\sec^2(x)}{\tan^2(x)}\times\dfrac{\tan(x)}{x}}{2\tan(x)\sec^2(x)} \\\\ M= \lim_{x\to0}\dfrac{\sin(x)\cos^3(x) - x\cos^2(x)}{2x\sin^2(x)}$

As $x\to0$, we get another 0/0 indeterminate form. Apply L'Hopital's rule again:

$\displaystyle M = \lim_{x\to0} \frac{\sin(x)\cos^3(x) - x\cos^2(x)}{2x\sin^2(x)} \\\\ M = \lim_{x\to0} \frac{\cos^4(x) - 3\sin^2(x)\cos^2(x) - \cos^2(x) + 2x\cos(x)\sin(x)}{2\sin^2(x)+4x\sin(x)\cos(x)}$

Recall the double angle identity for sin:

sin(2x) = 2 sin(x) cos(x)

Also, in the numerator we have

cos⁴(x) - cos²(x) = cos²(x) (cos²(x) - 1) = - cos²(x) sin²(x) = -1/4 sin²(2x)

So we can simplify M as

$\displaystyle M = \lim_{x\to0} \frac{x\sin(2x) - \sin^2(2x)}{2\sin^2(x)+2x\sin(2x)}$

This again yields 0/0. Apply L'Hopital's rule again:

$\displaystyle M = \lim_{x\to0} \frac{\sin(2x)+2x\cos(2x)-4\sin(2x)\cos(2x)}{2\sin(2x)+4x\cos(2x)+4\sin(x)\cos(x)} \\\\ M = \lim_{x\to0} \frac{\sin(2x) + 2x\cos(2x) - 2\sin(4x)}{4\sin(2x)+4x\cos(2x)}$

Once again, this gives 0/0. Apply L'Hopital's rule one last time:

$\displaystyle M = \lim_{x\to0}\frac{2\cos(2x)+2\cos(2x)-4x\sin(2x)-8\cos(4x)}{8\cos(2x)+4\cos(2x)-8x\sin(2x)} \\\\ M = \lim_{x\to0} \frac{4\cos(2x)-4x\sin(2x)-8\cos(4x)}{12\cos(2x)-8x\sin(2x)}$

Now as $x\to0$, the terms containing x and sin(nx) all go to 0, and we're left with

$M = \dfrac{4-8}{12} = -\dfrac13$

Then the original limit is

$L = \exp(M) = e^{-1/3} = \boxed{\dfrac1{\sqrt[3]{e}}}$