Prove that $\lim_{n \to \infty}{\sqrt{n+\sqrt{n}}} = \lim_{n \to \infty}{\sqrt{n}}$ This question is spurred by my discussion here: Find the value of : $\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}}}}-\sqrt{x}$
My question is how do you prove what @Ant is claiming? It looks like a very useful simplification  when the number of nested radical signs grow big. So I'm looking for a proof of:
$$\lim_{n \to \infty}{\sqrt{n+\sqrt{n}}} = \lim_{n \to \infty}{\sqrt{n}}$$
or at least a proof that one of the expressions can be substituted for the other in limit calculations. 
 A: Are you happy that 
$$\lim_{n \to \infty} \sqrt n = \infty?$$
If so you just need to note that
$$\sqrt n \le \sqrt{n + \sqrt n}$$
and then take the limit as $n \to \infty$.
A: Assuming you want to prove that $$\lim_{n\to\infty}\frac{\sqrt{n+\sqrt{n}}}{\sqrt{n}}=1$$ This follows by putting everything into one square root. $$\lim_{n\to\infty}\sqrt{\frac{n+\sqrt{n}}{n}}=\lim_{n\to\infty}\sqrt{1+\frac{1}{\sqrt{n}}}=1$$
Another way to show that is to show $$\sqrt{n}\leq\sqrt{n+\sqrt{n}}\leq\sqrt{n}+\frac12$$
A: Note however that: $$\lim_{n\to\infty} \left(\sqrt{n+\sqrt{n}}- \sqrt{n}\right)\ne 0$$
Multiply and divide by the rational conjugate:
$$
\require{cancel}
\begin{align}
\lim_{n\to\infty} \left(\sqrt{n+\sqrt{n}}- \sqrt{n}\right) & = \lim_{n\to\infty} \frac{\left(\sqrt{n+\sqrt{n}}- \sqrt{n}\right)\left(\sqrt{n+\sqrt{n}}+ \sqrt{n}\right)}{\sqrt{n+\sqrt{n}}+ \sqrt{n}} \\
 & = \lim_{n\to\infty} \frac{\cancel{n}+\sqrt{n} - \cancel{n}}{\sqrt{n+\sqrt{n}}+ \sqrt{n}} \\
 & = \lim_{n\to\infty} \frac{1}{\sqrt{1 + \cfrac{1}{\sqrt{n}}}+ 1} \\
 & = \frac{1}{2}
\end{align}
$$
