Validity of argument showing that $\frac{1}{\sqrt{n}+\frac{1}{\sqrt{n}}}\rightarrow\frac{1}{\sqrt{n}}\rightarrow0$ I was wondering if the following step is valid:
\begin{equation*}
\frac{1}{\sqrt{n}+\frac{1}{\sqrt{n}}}\rightarrow\frac{1}{\sqrt{n}}\rightarrow0
\end{equation*}
Since I'm saying
\begin{equation*}
\sqrt{n}+\frac{1}{\sqrt{n}}\rightarrow\sqrt{n}+0=\sqrt{n}
\end{equation*}
but not anything about $\sqrt{n}~~~(n\rightarrow\infty)$.
If it's not, is there a better way to do so?
 A: Since $$\sqrt{n}+\frac1{\sqrt{n}} \ge \sqrt{n} $$
We have $$\frac1{\sqrt{n}+\frac1{\sqrt{n}}} \le \frac1{\sqrt{n}} $$
Now you can use squeeze theorem.
A: Technically you can't do that. If you take the limit, all $n$ have to be dropped and replaced with the correct limits. However, in your head you can use this two-step approach, but if you write it down it's only one step
$$
\frac{1}{\sqrt{n}+\frac{1}{\sqrt{n}}} \to 0
$$
You can add for explanation that $\frac{1}{\sqrt{n}} \to 0$ and $\sqrt{n} \to \infty$. It depends where you need that, but usually you can use the fact that $\sqrt{n} \to \infty$ (otherwise you may have to show that as well).
A: Your argument is valid, albeit not rigorous. Imagine it this way. The given expression evaluates to:
$$\lim_{n\to \infty} \frac {\sqrt n}{n+1}$$
Which can be written as:
$$\lim_{n\to \infty}\frac {\sqrt n}{n} .\frac {n}{n+1}$$
Now, using algebra of limits:
$$\lim_{n\to \infty}\frac {\sqrt n}{n} .\lim_{n\to \infty}\frac {n}{n+1}$$
First limit tends to 0 while the second limit tends to 1, so the net limit would be 0, clearly.
