# Calculus. Finding limit of sequence. Trouble with L'Hopitals.

I am taking a calculus 2 class in the summer and I came across this problem.

" Find the limit of the sequence with the given nth term. $$a_n = \frac{2n}{\sqrt{n^2+1}}$$ "

I turned that into a limit form.

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

But I got stuck at that point because using L'Hopitals rule ended up seeming to repeat (more on this later).

After searching, I eventually understood the problem to be solved through pulling an $$n^2$$ out of the square root (turning it into n) and then simplifying and just applying the limit to get $$\frac{2}{1+0}$$ for a final answer of 2.

My question is why did using L'Hopital's rule not work?

The form after seems to be $$\infty/\infty$$, so L'Hopitals's should of worked. Here is my work.

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

At this point I applied the L'Hopitals's rule once more and arrived at the original limit. Did I make a mistake somewhere?

• "The form after seems to be $\infty/\infty$, so L'Hopitals's should [have] worked." - Unfortunately, just not true. Having the right form is a pre-condition for using L'Hopital's Rule, but does not guarantee that it will actually give you an answer. – David Jun 12 at 7:11
• Compute the limit of the square. – Claude Leibovici Jun 12 at 7:21

It did work, as you have $$\frac{2\sqrt{n^{2}+1}}{n}=2\sqrt{1+\frac{1}{n^{2}}}\rightarrow2$$ the same as the original limit. The fact that it doesn't give you a seemingly simpler answer is not a problem.
• @MusaabAli Yes to your second statement. I wouldn't call those forms indeterminate though, as you can easily calculate both by pulling $n^{2}$ out of the square root. – eranreches Jun 12 at 8:24
$$\frac{2n}{\sqrt{n^2+1}}=\frac{2n}{|n|\sqrt{1+\frac{1}{n^2}}}=\frac{2}{\sqrt{1+\frac{1}{n^2}}}$$
Last equality follows from $$n>0$$.