Limit at infinity involving $e$ I am to find the limit of $$\lim_{x \to \infty} \left(1+\frac{x}{5x^3+x^2+8}\right)^ {\dfrac{x^3+8}{x}}$$ 
I could not find the proper substitution here. I would be happy if someone could shed some light. Thanks.

$$\lim_{x\to\infty}\left(1+\frac{x}{5x^3+x^2+8}\right)^ {x^2+8/x}=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^ {x^2+8/x}=$$
$$=\lim_{x\to\infty}\left(\left(1+\frac{1}{5x^2+x+8/x}\right)^{5x^2+x+8/x} \right)^{\frac{x^2+8/x}{5x^2+x+8/x}}=e^{1/5}$$
because
$$\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{5x^2+x+8/x}=e$$
and
$$\lim_{x\to\infty}{\frac{x^2+8/x}{5x^2+x+8/x}}=\frac{1}{5}$$
