Using definition of limit to prove limit

I am having trouble with this problem:

Use the definition of limit to prove that:
$$\lim_{x\to \infty} \frac{\sin x}{x(\sin x)^2 +1} = \,?$$

I have concluded that the limit must be 0, but I am having trouble proving it.

Using the limit definition, I must show that

$$\left\lvert \frac{\sin x}{x(\sin x)^2 +1}- 0\right\rvert < \epsilon$$

Answer

Clearly we may assume that $x>0$. Consider two cases:

• if $|\sqrt x\sin x|<1$ then $|\sqrt x\sin x|
  
• if $|\sqrt x\sin x|\ge1$ then $|\sqrt x\sin x|\le x\sin^2x
  

Hence in all cases we have
$$\Bigl|\frac{\sin x}{x\sin^2x+1}\Bigr|<\frac1{\sqrt x}\ ;$$
so given $\varepsilon>0$, taking $x>1/\varepsilon^2$ guarantees that the LHS is less than $\varepsilon$. I'll leave you to turn this into a formal proof.