Limit of Exponential Function I am stuck on a question involving the limit of an exponential function, as follows
$$\lim_{z \to \infty} \left ( 1-\frac{4}{z+3} \right )^{z-2}$$
The following hint is given:
$$ \text{Assume that} \qquad \lim_{x \to 0}\left ( \frac{\ln(1+x)}{x} \right )=1$$
My first thought was to address the behaviour of the function within the brackets: $$\lim_{z \to \infty} \left ( 1-\frac{4}{z+3} \right )=1$$ Of course $1^{z-2}$ as $ z \to \infty$ is equal to one. I don't think this is correct. Wolfram Alpha informs me $$\lim_{z \to \infty} \left ( 1-\frac{4}{z+3} \right )^{z-2}=\frac{1}{e^{4}},$$ and I have not used the given hint. This is my first time solving a limit involving an exponent with a variable - I am missing something. Thank you for any help.

If you are allowed to use the fact that 
$$\lim_{y \to \infty}\left(1+\frac{a}{y}\right)^y=e^a,$$
then you can rewrite your expression as
$$\lim_{x\to \infty}\left(\left(1-\frac{4}{x+3}\right)^{x+3}\right)^{\frac{x-2}{x+3}}.$$
As $x\to \infty$, $x+3\to \infty$, and therefore
$$\lim_{x\to \infty}\left(1-\frac{4}{x+3}\right)^{x+3}=e^{-4}.$$
But 
$$\lim_{x\to\infty} \frac{x-2}{x+3}=1,$$
and the result follows.
