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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?

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    $\begingroup$ Note that this limit is just $$\frac1{x^2+1}\cdot\lim_{y\to0}\frac{\sin{(y)}}y=\frac1{x^2+1}$$where $y=x/n\to0$. $\endgroup$ Jun 9, 2020 at 17:21

2 Answers 2

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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.

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  • $\begingroup$ I am not sure I understand. Can you break it down a bit please? $\endgroup$
    – Lauren Sin
    Jun 9, 2020 at 18:08
  • $\begingroup$ $\lim_n n \sin(x/n) =x \lim_{\theta\to 0^+}\sin(\theta)/\theta =x$ $\endgroup$
    – Integrand
    Jun 9, 2020 at 18:09
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$$\lim_{n \to \infty} \sin{\left(\frac{x}{n}\right)} \approx \frac{x}{n}$$ Therefore, $$\lim_{n \to \infty} \frac{n \cdot \frac{x}{n}}{x(x^2+1)}=\frac{1}{x^2+1}$$

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