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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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    @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.2012-02-23

2 Answers 2

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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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    @kellax: right, that's exactly what I wrote in my answer - $f$urther you have that limits of both numerator and denominator exist, and the latter is non-zero. Hence you can simply divide limits2012-02-23
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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 $