Find $\lim_{x\rightarrow -\infty } \frac{3x^{2}-\sin(5x)}{x^{2}+2}$ 
$\lim_{x\rightarrow -\infty } \frac{3x^{2}-\sin(5x)}{x^{2}+2}$
A. 0 
B. 1 
C. 2 
D. 3

After using L'hopital's rule again and again I got this expression:
$$\lim_{x\rightarrow -\infty } \frac{6+25\sin(5x)}{2}$$
But how do we proceed, what is the value of $\sin(-\infty)$?
Any help will be appreciated!
 A: Application of L'hopitals rule for the first time gives,
$$\lim_{x \to -\infty} \frac{6x-5\cos (5x)}{2x}$$
We can try to do it again but it won't be useful $\lim_{x \to -\infty} \sin (x)$ does not exist because it oscillates between $-1$ and $1$ for $\frac{\pi}{2}+2\pi k$ and $\frac{3\pi}{2}+2\pi k$ so L'hopitals rule will not give a sensible answer.
Instead  we may try more elementary approaches and write is as follows.
$$\lim_{x \to -\infty} \frac{6-5\frac{\cos(5x)}{x}}{2}$$
At this point note that the quaintly $\frac{\cos (5x)}{x}$ goes to zero because the top is bounded by $-1$ and $1$.
So the answer is $3$.
A: You can also solve by the Squeeze Theorem
$$\lim _{x\to \:-\infty \:}\left(\frac{3x^2-\left(-1\right)}{x^2+2}\right)\le \lim \:_{x\to \:-\infty \:}\left(\frac{3x^2-\sin \left(5x\right)}{x^2+2}\right)\le \lim \:_{x\to \:-\infty \:}\left(\frac{3x^2-1}{x^2+2}\right)$$
$$\color{red}{3}\le \lim \:_{x\to \:-\infty \:}\left(\frac{3x^2-\sin \left(5x\right)}{x^2+2}\right)\le \color{red}{3}$$
