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.

Note that
$$\left(1+\frac{2}{x}+\frac{3}{x^2}\right)^{7x}=\left[\left(1+ \frac{2}{x}+\frac{3}{x^2}\right)^{\frac1{\left(\frac{2}{x}+\frac{3}{x^2}\right)}}\right]^{{7x}{\left(\frac{2}{x}+\frac{3}{x^2}\right)}}\to e^{14}$$
