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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:

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

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    $\begingroup$ I do not think you can apply L'Hospital three times here $\endgroup$ – Vasya Aug 29 at 16:51
  • $\begingroup$ Can you please explain why? $\endgroup$ – Avijit Dasgupta Aug 29 at 16:52
  • $\begingroup$ since $|sin(x)|\le 1$ the dominant term would be only the $3x^2$ and $x^2$ in the nominator and denominator, and therefore the limit result is $3$ $\endgroup$ – kvphxga Aug 29 at 16:52
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    $\begingroup$ You can apply L'Hospital if both numerator and denominator tend to either zero or infinity so what happens after you apply it two times? $\endgroup$ – Vasya Aug 29 at 16:54
  • $\begingroup$ @Vasya You can't even apply it two times because the limit produced doesn't exist in the second application. $\endgroup$ – Peter Foreman Aug 29 at 16:56
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Hint: Write $$\frac{3+\frac{\sin(x)}{x^2}}{1+\frac{\sin^2(x)}{x^2}}$$

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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 $$ \frac{3x^2+\sin x}{x^2+\sin^2x}\sim_\infty \frac{3\not x^2}{\not x^2}=3. $$

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