What is the solution to the following limit?

Question

How to solve the following limit problem?

$$\lim_{x\to\infty} \frac{3x^2+\sin x}{x^2+(\sin x)^2}$$. I have tried the following and arrived at a solution. But not sure if it is correct.

Applying L'Hospital's rule taking derivative thrice I get the following expression:

$$\lim_{x\to\infty} \frac{\cos x}{8\cos 2x} = 0$$

Answer 1

Hint: Write $$\frac{3+\frac{\sin(x)}{x^2}}{1+\frac{\sin^2(x)}{x^2}}$$

Answer 2

Use asymptotic equivalence: as a sine is bounded, $3x^2+\sin x\sim_\infty 3x^2$, $x^2+\sin^2x\sim_\infty x^2$, so $$ \frac{3x^2+\sin x}{x^2+\sin^2x}\sim_\infty \frac{3\not x^2}{\not x^2}=3. $$