Finding $\lim_{x\to \infty}(1+\frac{2}{x}+\frac{3}{x^2})^{7x}$ I am having some diffuculty finding the limit for this expression and would appreciate if anyone could give a hint, as to how to continue. I know the limit must be $e^{14}$ (trough an engine) and I can show it for 
$(1+\frac{2}{x}+\frac{1}{x^2})^{7x}$ like 
$\lim_{x\to \infty}(1+\frac{2}{x}+\frac{1}{x^2})^{7x} = ((1+\frac{1}{x})^2)^{7x} = (1+\frac{1}{x})^{x \cdot 7 \cdot 2} = e^{7 \cdot 2} = e^{14}$
However the $\frac{3}{x^2}$ in $(1+\frac{2}{x}+\frac{3}{x^2})^{7x}$ is causing problems for me I have:
$\lim_{x\to \infty}(1+\frac{2}{x}+\frac{3}{x^2})^{7x} = (1+\frac{2}{x}+\frac{1}{x^2} +\frac{2}{x^2})^{7x} = ((1+\frac{1}{x})^2+\frac{2}{x^2})^{7x} =...$
but I'm not sure how to continue (how to get rid of $\frac{2}{x^2}$) I was thinking that using the binomial theorem might somehow reduce the$\frac{2}{x^2}$ for $\lim_{x\to \infty}$, but I'm not sure how to do that for an x in the exponent, or if it is useful, or if it is even allowed in this case.
If anyone could point me in the right direction, as to how to continue I would be very grateful.

It can be convenient to use Taylor series when doing problems like this:
$$\lim_{x \to \infty} \left(1 + \frac{2}{x} + \frac{3}{x^2}\right)^{7x} = \lim_{x \to \infty} e^{7x\ln(1+\frac{2}{x}+\frac{3}{x^2})} = \lim_{x\to \infty} e^{7x\left( \frac{2}{x} + \frac{3}{x^2} + \mathcal{O}\left(\left((\frac{2}{x} + \frac{3}{x^2}\right)^2\right) \right)} = \lim_{x\to \infty} e^{14 + \frac{21}{x}} = e^{14}$$
