Finding: $\lim\limits_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$ 
I'm running into problems with this limit:
  $$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$$

I've tried using l'Hospitals rule, however we will alway keep the $\cos(\pi x)$ expression, as well for $\sin(5\pi x)$. Also, terms do not cancel out with $\sin$ and $\pi$ since the product with $5$. 
Can anyone give me a hint solving this?
Thanks in advance
Kind regards,
 A: Hint: The functions sine and cosine are bounded. Therefore, even though they can oscillate wildly at infinity, they can be controlled. 
This implies that in limit operations, we have the following rules of thumb that are correct if used appropriately:
$$1.\lim_{x\to \infty}(0 \times \sin(x)) = 0$$
$$2.\lim_{x\to \infty}(0 \times \cos(x)) = 0$$
$$3.\lim_{x\to \infty}(\frac{\sin(x)}{\infty}) = 0$$
$$4.\lim_{x\to \infty}(\frac{\cos(x)}{\infty}) = 0$$
In all of the above equalities, $0$ in the parentheses is meant to be taken as something infinitesimal. Note that $3.$ and $4.$ can be thought as special cases of $1.$ and $2.$ respectively.
Now factor out $x^2$ in the numerator and $x^4$ under the square root in the denominator and proceed.
If that hint is not enough, hover your mouse over the orange area:

 $$\lim_{x \to -\infty}\frac{6x^2+5\cos\pi x}{\sqrt{x^4+5\sin 5\pi x}} = \lim_{x \to -\infty}\frac{x^2\left(6 + \frac{5\cos \pi x}{x^2}\right)}{\sqrt{x^4(1 + \frac{5\sin 5\pi x}{x^4})}}=\lim_{x \to -\infty}\frac{6 + \frac{5\cos \pi x}{x^2}}{\sqrt{1 + \frac{5\sin 5\pi x}{x^4}}}=6$$

A: Easy using Squeeze theorem
$$ 6x^2-5\le6x^2+5\cos{\pi x}\le 6x^2+5$$
and 
$$ 6x^4-1\le6x^2+\sin{5\pi x}\le 6x^4+1$$
then for $x<-1$ we have 
$$\frac{6x^2-5}{\sqrt{x^4+1}}\le \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}} \le \frac{6x^2+5}{\sqrt{x^4-1}}$$ 
By squeeze theorem we get
$$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}} =6$$
A: Writing out like
$$\frac{6x^2+5\cos\pi x}{\sqrt{x^4+5\sin 5\pi x}} = \frac{x^2\left(6 + \frac{5\cos \pi x}{x^2}\right)}{x^2\sqrt{1 + \frac{5\sin 5\pi x}{x^4}}}$$
should do the trick.
A: With equivalents:
Since the trigonometric functions are bounded, $6x^2+5\cos \pi x\sim_\infty 6x^2$, $\;x^4+5\sin 5\pi x\sim_\infty x^4$, so
$$\frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}\sim_\infty \frac{6x^2}{\sqrt{x^4}}=6.$$
