# Limit of a sequence

How can I show that $\lim\limits_{x \rightarrow \infty} (1+ \frac{1}{4x^2 + 8/x})^{x^2+8/x} =e^\frac{1}{4}$ using that $\lim\limits_{x \rightarrow \infty} (1+\frac{1}{x} ) ^x = e$ ?

I can obviously take out a 4 from the denominator, but how can I get the same power as what I'll get in the denominator?

Try showing $\lim_{x\rightarrow \infty}(1+a/x)^x=e^a$. Hint: you know that $\lim_{x\rightarrow \infty}(1+a/x)^{x/a}=e$ –  Cantor Nov 24 '12 at 18:29
HINT: $$\left(1+ \dfrac{1}{4x^2 + \dfrac{8}{x}}\right)^{x^2+\dfrac{8}{x}} = \left[\left(1+ \dfrac{1}{4\left(x^2 + \dfrac{2}{x}\right)}\right)^{4\left(x^2+\dfrac{2}{x}\right)}\right]^\dfrac{1}{4}\left(1+ \dfrac{1}{4x^2 + \dfrac{8}{x}}\right)^{\dfrac{6}{x}}.$$