0
$\begingroup$

How to find limit of $$\lim_{n\to \infty} \frac{n \sin\frac{x}{n}}{x(x^2+1)}$$ without L'Hospital's rule?

I thought rewriting $\sin(\frac{x}{n})$ using Taylor expansion would work but it didn't help. I solved it using L'Hospital's rule: $$\lim_{n\to \infty} \frac{n \sin\frac{x}{n}}{x(x^2+1)} = \lim_{n\to \infty} \frac{ \sin\frac{x}{n}}{(x^2+1)\frac{x}{n}}=\lim_{n\to \infty} \frac{ \cos\frac{x}{n}}{x^2+1}=\frac{1}{x^2+1},$$

but I would like to know how to solve it without the rule. If it is not possible without L'Hospital's rule is there any other way to solve the limit with the rule apart from what I've written?

$\endgroup$
  • 3
    $\begingroup$ Note that this limit is just $$\frac1{x^2+1}\cdot\lim_{y\to0}\frac{\sin{(y)}}y=\frac1{x^2+1}$$where $y=x/n\to0$. $\endgroup$ – Peter Foreman Jun 9 '20 at 17:21
1
$\begingroup$

For $\theta$ near $0$, we have $$ \cos(\theta)\leq \sin(\theta)/\theta \leq 1 $$So with $\theta=x/n$, as $n$ gets large $n\sin(x/n)$ approaches $x$, whence you recover the original result.

$\endgroup$
  • $\begingroup$ I am not sure I understand. Can you break it down a bit please? $\endgroup$ – Lauren Sin Jun 9 '20 at 18:08
  • $\begingroup$ $\lim_n n \sin(x/n) =x \lim_{\theta\to 0^+}\sin(\theta)/\theta =x$ $\endgroup$ – FearfulSymmetry Jun 9 '20 at 18:09
1
$\begingroup$

$$\lim_{n \to \infty} \sin{\left(\frac{x}{n}\right)} \approx \frac{x}{n}$$ Therefore, $$\lim_{n \to \infty} \frac{n \cdot \frac{x}{n}}{x(x^2+1)}=\frac{1}{x^2+1}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.