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}} $