Is it possible to apply L'Hospital to $ \lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8}))) $?

Question

I have a question about how to calculate this limit; can I apply L'Hospital?

$$ \lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8}))) $$

Is it possible to make a change of variable such as $$ t^2 = \frac{1}{x^4}\;?$$

Answer 1

The way to go is $|\sin x |\leq |x|$. Therefore, $\sin^2\left(\frac{2}{x^4}\right)\leq \frac{4}{x^8}$. It should be clear from here how to use Squeeze Theorem. L'Hospitals was undoubtedly not the intent of the problem.

Answer 2

$$\lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8})))$$

$$ \lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} ) = \lim_{x \to\infty}\left(\dfrac{1}{e^{x^2}}+\dfrac{2}{x} \right) = 0$$

$$ \lim_{x \to\infty}\sin^2\left(\frac{2}{x^4}\right)x^8 = \left[ 2\lim_{x \to\infty} \dfrac{\sin\left(\frac{2}{x^4}\right)}{\frac{2}{x^4}}\right]^2 = 4$$

$$ \lim_{x \to\infty} \cos \left(\frac{3}{x^8}\right) = 1 $$