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.
 A: 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.
A: This is what you can do. first let $z-2=n$. then your expression reduces to $\left(1-\frac{4}{n+5}\right)^n$. Now simplify to get $\left(\frac{n+1}{n+5}\right)^n$ and make another substitution: $x=1/n$. Then you get $$\left(\frac{x+1}{1+5x}\right)^{1/x}.$$ Note that at this point you take limit as $x$ goes to zero since you had limit as $n$ goes to infinity. take the natural log of this expression and bring the power down so that you make use of the hint given and find the the exponential to get the the answer.
A: Taking logs you need to find
$$\lim_{z \rightarrow \infty}(z-2)\log(1 - {4 \over z + 3})$$
You will then exponentiate the result. You can rewrite the above as 
$$\lim_{z \rightarrow \infty}(z-2)\bigg({-{4 \over z+ 3}}\bigg){\log(1 - {4 \over z + 3}) \over {-{4 \over z + 3}}}$$
As $z \rightarrow \infty$, ${\displaystyle {4 \over z + 3} \rightarrow 0}$, so...
