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.