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.


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?

  • $\begingroup$ "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. $\endgroup$ – David Jun 12 at 7:11
  • $\begingroup$ Compute the limit of the square. $\endgroup$ – 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.

  • $\begingroup$ So the issue with my problem solving wasn't that L'Hopital's rule was giving multiple indeterminate forms, but that L'Hopital's rule was not needed in the first place? $\endgroup$ – Musaab Ali Jun 12 at 8:18
  • $\begingroup$ @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. $\endgroup$ – eranreches Jun 12 at 8:24

Hint (without L'hopital):


Last equality follows from $n>0$.


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