# 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

$$\sqrt{n^2+1} = n \sqrt{1+\frac{1}{n^2}}$$

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You LaTeX is very nice, but please use $\cdot$ \cdot instead of $*$ which may be confusing. –  Ilya Feb 23 '12 at 14:44
You have a sequence, not a series. –  David Mitra Feb 23 '12 at 14:56
@kellax: If anywhere in your work, you have an explicit expression where you are dividing by $0$, and even worse, dividing $0$ by $0$, there is a mistake. Explicit division by $0$ occurs four times in the second line of your displayed equation. –  André Nicolas Feb 23 '12 at 15:26

I didn't get how it appeared $\sqrt{n^2+\frac1n}$ already in the first numerator - but the answer you got is correct. Nevertheless, for such limits with $n\to\infty$ and powers both in numerator and denominator, you should always divide both numerator and denominator by the highest power of $n$. Here the highest power is $3/2$ so you get $$\lim\limits_n\frac{n\sqrt{n}+n}{\sqrt{n^3}+2} = \lim\limits_n\frac{1+\frac{1}{\sqrt{n}}}{{1+\frac{2}{n^{3/2}}}} = 1$$ since both limits $\lim\limits_n\left(1+\frac{1}{\sqrt{n}}\right)$ and $\lim\limits_n\left(1+\frac{2}{n^{3/2}}\right)$ exist, finite and equal to the same number: $1$.

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I ment $$\sqrt{n^2 *\frac{n}{n^2}}$$ –  Sterling Duchess Feb 23 '12 at 14:42
@kellax I see - anyway, your way of solution is not very clear and it might be informal (e.g. taking $\sqrt{0}$ inside the limit) - so I would advise you just to divide both numerator and denominator by the highest power of $n$, that is a simple and very usual method. –  Ilya Feb 23 '12 at 15:05
In this case i would devide both numerator and denominator expressions by $$\sqrt{n^3}$$ and that would give me $$\frac{(1 + \frac{1}{\sqrt{n}})}{(1+\frac{2}{n^{\frac{3}{2}}})}$$ –  Sterling Duchess Feb 23 '12 at 15:26
@kellax: right, that's exactly what I wrote in my answer - further you have that limits of both numerator and denominator exist, and the latter is non-zero. Hence you can simply divide limits –  Ilya Feb 23 '12 at 15:31

A more readable form would be

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

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