To compute the following limit problem :
$$\lim_{n\to\infty} \sqrt[n] {\frac {1 - n^2}{1 + n^2}}$$.
Why it's not mathmatically correct way to compute the limit by making the manipulation : $n = \frac{1}{t}$ and compute the limit in the following manner, $n \rightarrow \infty$ so, $t \to 0$?
What's the correct way?
Denote
$$L=\lim_{n\to\infty} \sqrt[n] {\frac {1 - n^2}{1 + n^2}}$$
$$\ln L=\lim_{n\to\infty} \frac 1n \ln\Bigg({\frac {1 - n^2}{1 + n^2}}\Bigg)=\frac{i\pi}{\to\infty}\to0$$
$$L=e^0=1$$
$$L=\lim_{n\to\infty} \sqrt[n] {\frac {1 - n^2}{1 + n^2}}$$ Applying Cauchy-d'Alembert criterion: $$L=\lim_{n\to\infty} {\frac {1 - (n+1)^2}{1 + (n+1)^2}}{\frac {1 + n^2}{1 - n^2}}$$ $$L=\lim_{n\to\infty} {\frac {n^2+2n}{n^2+2n+2}}{\frac { n^2+1}{ n^2-1}}$$ $$\implies L=1$$
As the natural numbers are a subset of the real numbers, $\lim_{n\to\infty} \sqrt[n] {\frac {1 - n^2}{1 + n^2}} = \lim_{u \to\infty} \sqrt[u] {\frac {1 - u^2}{1 + u^2}}$, where $u$ is a real number.
Let $u = \tan x, x \in [0, \frac{\pi}{2})$ and $x \in \mathbb R$, as the range of $\tan x$ is $[0, \infty)$ in the given domain.
Then $1 - u^2 = \frac{\cos^2 x - \sin^2 x}{\cos^2 x} = \cos(2x) \sec^2 x$, and $1 + u^2 = \sec^2 x$. Hence we need to find:
$$\lim_{x \to \pi/2} {\cos(2x)}^{1 / \tan x} =\lim_{x \to \pi/2} (\cos 2x)^0 = \boxed{1}.$$
as $\lim_{x \to \pi/2} \tan x = +\infty \Rightarrow \lim_{x \to \pi/2} \frac{1}{\tan x} = 0$.
