<s>[INST] 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,
 [/INST] 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
$$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}} =6$$
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