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}))) $? 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 $$ t^2 = \frac{1}{x^4}\;?$$

$$\lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )\sin^2(\frac{2}{x^4})x^8\cos(\frac{3}{x^8})))$$
$$    \lim_{x \to\infty}(e^{-x^2}+\frac{2}{x} )
   = \lim_{x \to\infty}\left(\dfrac{1}{e^{x^2}}+\dfrac{2}{x} \right)
   = 0$$
$$   \lim_{x \to\infty}\sin^2\left(\frac{2}{x^4}\right)x^8
   = \left[ 2\lim_{x \to\infty}
           \dfrac{\sin\left(\frac{2}{x^4}\right)}{\frac{2}{x^4}}\right]^2
   = 4$$
$$   
      \lim_{x \to\infty} \cos \left(\frac{3}{x^8}\right) = 1
$$
