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

Question

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?

Answer 1

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.

Answer 2

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