Proving a limit exists using the definition of a limit $ \lim_{x\to\infty} {\sqrt{f(x)+1}}) = {\sqrt{L+1}} $ Let $f(x)$ be a function such that $f(x)\geq -1$ for every $x$. Suppose, $\lim_{x\to\infty} f(x) = L$ and that $L\leq-1$. Prove by using the definition of a limit that:
$  \lim_{x\to\infty} {\sqrt{f(x)+1}}) = {\sqrt{L+1}} $
Hi I would very much appreciate an explanation on how to solve this problem using the definition. I know the definitions well but don't succeed on solving accurately, maybe some tips in general of how to solve these kind of problems. Thank you!

Supposing $L$ is finite and $L\geq -1$: $\displaystyle\lim_{x\to\infty}f(x)=L$ means that (by definition) for every given $\epsilon>0$, there exists a real number $N_1$ (depending on $\epsilon)$ such that the inequality $|f(x)-L|<\epsilon$ is satisfied for all $x\geq N_1$. Also for the same epsilon there is $N_2$  such that $|f(x)-L|<\sqrt \epsilon$  for all $x\geq N_2$. Now, look at $$\bigg|\sqrt{f(x)+1}-\sqrt{L+1}\big|^2=\big|\sqrt{f(x)+1}-\sqrt{L+1}\big|\big|\sqrt{f(x)+1}-\sqrt{L+1}\big|$$
$$\leq \big|\sqrt{f(x)+1}-\sqrt{L+1}\big|\big|\sqrt{f(x)+1}+\sqrt{L+1}\big|=|f(x)-L|^2<\epsilon\quad (\text{by difference of squares}).$$
Thus if we let $N=\max\{N_1, N_2\}$, then we see that for the given $\epsilon>0$
the inequality $$\big|\sqrt{f(x)+1}+\sqrt{L+1}\big|<\sqrt\epsilon$$
is satisfied for all $x\geq N$,
where we have also used the fact that $|a-b|\leq |a+b|$ for $a,\,b\geq 0.$ This shows that $\displaystyle\lim_{ x\to \infty}\sqrt{f(x)+1}=\sqrt{L+1}$.
If $L$ were $\infty$, we would use this definition:
$\displaystyle\lim_{x\to\infty}f(x)=\infty\Longrightarrow$ for every  $\epsilon>0$, there exists a real number $N$ (depending on $\epsilon)$ such that the inequality $f(x)\geq \epsilon$ is satisfied for all $x\geq N.$ In this case using almost similar idea you will show that $\displaystyle\lim_{x\to\infty}\sqrt{f(x)+1}=\infty.$
