Question

`\lim_(n \to \0)(x/(tan⁡(x))^(cot⁡(x)^2 )`

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((x)/(\tan(x))\right)^(\cot^2(x)) = \exp\left(\cot^2(x) \ln\left((x)/(\tan(x))\right)\right) = \exp\left((\ln\left((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((\ln\left((x)/(\tan(x))\right))/(\tan^2(x))\right) = \exp\left(\lim_(x\to0)(\ln\left((x)/(\tan(x))\right))/(\tan^2(x))\right)`

Let M denote the remaining limit.

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

`\displaystyle M = \lim_(x\to0)(\ln\left((x)/(\tan(x))\right))/(\tan^2(x)) \\\\ M = \lim_(x\to0)((\tan(x)-x\sec^2(x))/(\tan^2(x))*(\tan(x))/(x))/(2\tan(x)\sec^2(x))`

Simplify and rewrite this in terms of sin and cos :

`\displaystyle M = \lim_(x\to0) ((\tan(x)-x\sec^2(x))/(\tan^2(x))*(\tan(x))/(x))/(2\tan(x)\sec^2(x)) \\\\ M= \lim_(x\to0)(\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) (\sin(x)\cos^3(x) - x\cos^2(x))/(2x\sin^2(x)) \\\\ M = \lim_(x\to0) (\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) (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) (\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) (\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)(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) (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 = (4-8)/(12) = -\frac13`

Then the original limit is

`L = \exp(M) = e^(-1/3) = \boxed{\frac1{\sqrt[3]{e}}}`