# How to calculate the limit $\lim_{n\to\infty} \frac{\sqrt{n+1} (1+n^2)}{\sqrt{n} (2 +n^2+2n)}$? Is my reasoning correct?

$$\lim_{n\to\infty} \frac{\sqrt{n+1} (1+n^2)}{\sqrt{n} (2 +n^2+2n)}$$

The answer to this is $1$ and I'm not fully sure why. If we divide each term by the highest power of $n$ in the denominator (which is $n$? since $n^{\frac{1}{2}}\times n^2$ will give us $n^1$?) I'm just not seeing how to get the answer is $1$. This is a question related to the ratio test, which is fine, but the limit is throwing me. Thank you for any help.

No, your reasoning is a little bit off - actually we have $$n^{1/2}n=n^{3/2}$$ but this is not really the problem here - you can't apply this here since in the numerator we have $\sqrt{n+1}\cdot n^2$, so you can't simply add the exponents since the arguments don't match.

I'd suggest the following approach: $$\lim_{n\to\infty} \frac{\sqrt{n+1} (1+n^2)}{\sqrt{n} (2 +n^2+2n)}=\lim_{n\to\infty} \frac{\sqrt{n+1}}{\sqrt{n}}\lim_{n\to\infty}\frac{ (1+n^2)}{(2 +n^2+2n)}$$ This holds, since both limits do exist (which we have to prove!). The first one is $$\lim_{n\to\infty} \frac{\sqrt{n+1}}{\sqrt{n}}=\sqrt{\lim_{n\to\infty} \frac{{n+1}}{{n}}}=1$$ and the second one $$\lim_{n\to\infty}\frac{ 1+n^2}{2 +n^2+2n}=\lim_{n\to\infty}\frac{ n^2(1+1/n^2)}{n^2(2/n^2 +1+2/n)}=1$$ So we have $$\lim_{n\to\infty} \frac{\sqrt{n+1} (1+n^2)}{\sqrt{n} (2 +n^2+2n)}=\lim_{n\to\infty} \frac{\sqrt{n+1}}{\sqrt{n}}\lim_{n\to\infty}\frac{ (1+n^2)}{(2 +n^2+2n)}=1\times1=1$$

• Thank you very much, very well explained and I understand now. – moony Nov 27 '17 at 15:55

$n^{1/2}\times n^2$ is $n^{2.5}$ - you add the exponents when you multiply two powers. So the fraction reduces to $$\frac{\sqrt{1+n^{-1}}(1+n^{-2})}{1+2n^{-2}+2n^{-1}}$$ and both top and bottom tend to $1$.

Divide both the numerator and the denominator by $n^2\sqrt n$. After you've done that, the limits of both the numerator and the denominator will both be equal to $1$.

Hint: you can rewrite it this way:

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

Define $a_n\sim b_n$ if $$a_n/b_n\to1$$ as $n\to\infty$. Note that $$\sqrt{n+1} (1+n^2)\sim n^2\sqrt{n}$$ while $$\sqrt{n} (2 +n^2+2n)\sim n^2\sqrt{n}.$$ Hence $$\frac{\sqrt{n+1} (1+n^2)}{\sqrt{n} (2 +n^2+2n)}\to1$$ as $n\to\infty$. We have used the fact that if $a_n\sim b_n$ and $c_n\sim d_n$, then $a_n/c_n\sim b_n/d_n$ since $$\frac{a_n/c_n}{b_n/d_n}=\frac{a_n}{b_n}\frac{d_n}{c_n}\to1.$$