Finding a limit of a series How would you calculate this limit it just blew me off on my midterms i seem to have calculated the limit correctly but my process is bougus < what my friend said.
$$
\lim_{n\to\infty}\frac{n \sqrt{n} +n}{\sqrt{n^3}+2}
$$
How i calculated the limit:
$$
\lim_{n\to\infty}\frac{n \sqrt{n^2  \frac{n}{n^2}{}} +n}{\sqrt{n^4  n^{-1}}+2} =
\lim_{n\to\infty}\frac{n^2 \sqrt{\frac{n}{n^2}{}} +n}{n^2\sqrt{n^2 n^{-1}}+2} =
\lim_{n\to\infty}\frac{n^2 \sqrt{\frac{n}{n^2}{}} +n}{n^2\sqrt{n^2 n^{-1}}+2}\\

\frac{n^2 \sqrt{\frac{1}{n}{}} +n}{n^2\sqrt{n^2 n^{-1}}+2}=
\frac{\frac{n^2\sqrt{0}}{n^2\sqrt{0}}+\frac{n}{n^2\sqrt{0}}}{\frac{n^2\sqrt{0}}{n^2\sqrt{0}}+\frac{2}{n^2\sqrt{0}}}= \frac{1}{1}=1
$$
I followed a book where an example was given where it said you can transform
a expression like so:
$$
\frac{1}{n} \sqrt{n^2 + 2} = \sqrt{\frac{1}{n^2} (n^2+2)}
$$
or
$$
\sqrt{n^2+1} = n \sqrt{1+\frac{1}{n^2}}
$$

A more readable form would be
$$
\large {
\lim_{n\to\infty}\frac{n \sqrt{n} +n}{\sqrt{n^3}+2} \hspace{4pt} = \hspace{4pt} \lim_{n\to\infty}\frac{n^{\frac{3}{2}}+n}{n^{\frac{3}{2}}+2} 

\hspace{4pt} = \hspace{4pt} \lim_{n\to\infty} 
\frac
{1+
\frac{1}{\sqrt{n}}
}
{1+\frac{2}
{n^{\frac{3}{2}}}
}
} \hspace{4pt} = \hspace{4pt} 1
$$
