Use arithmetic of limits to find $\lim_{n\to\infty} \frac{\sqrt{n}-1}{\sqrt{n}+1}$ $\lim_{n\to\infty} \frac{\sqrt{n}-1}{\sqrt{n}+1}$
Since $\lim_{x\to\infty} \sqrt{n}-1=\infty$ and $\lim_{n\to\infty}\sqrt{n}+1=\infty$
$\implies \lim_{n\to\infty} \frac{\sqrt{n}-1}{\sqrt{n}+1} = \lim_{n\to\infty} \frac{\infty}{\infty}=1$?
Thanks.
 A: write your Limit in the form $$\frac{1-\frac{1}{\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}$$
A: Dr. Sonnhard Graubner gave a good hint and I give another approach below, but it might even be more important for you to realize why your method fails. By the same reasoning, you would say:
$$\lim_{n\to\infty} \frac{n+1}{n^2}=\color{red}{1}$$
because $\lim_{n\to\infty} \left( n+1 \right)= \infty$ and $\lim_{n\to\infty} n^2= \infty$, while clearly:
$$\lim_{n\to\infty} \frac{n+1}{n^2}=\lim_{n\to\infty} \left( \frac{1}{n}+\frac{1}{n^2} \right) = 0$$
You cannot simplify $\frac{\infty}{\infty}$ to $1$, it is a so called indeterminate form.

Alternative:
$$\frac{\sqrt{n}\color{blue}{-1}}{\sqrt{n}+1}=\frac{\sqrt{n}\color{blue}{+1-2}}{\sqrt{n}+1} = \frac{\sqrt{n}\color{blue}{+1}}{\sqrt{n}+1}-\frac{\color{blue}{2}}{\sqrt{n}+1}= 1-\frac{2}{\sqrt{n}+1}$$
So:
$$\lim_{n\to\infty} \frac{\sqrt{n}-1}{\sqrt{n}+1}= \lim_{n\to\infty} \left( 1-\frac{2}{\sqrt{n}+1} \right)= 1- \underbrace{\lim_{n\to\infty} \frac{2}{\sqrt{n}+1}}_{\to 0}=1$$
A: By substitution : 
Set $t=\sqrt n$. It comes down to finding the limit as $t$ tends to $\infty$ of the rational function $\;\dfrac{t-1}{t+1}$, for which there's a high-school theorem: it's the limit of the ratio of the leading terms. This happens to be  $1$.
