For what values of $k$ does this limit exist?
$$\lim_{x \to-\infty} \frac{5^{kx}-1}{5^{-2x} + 1}$$
Progress: I worked it by dividing everything by $5^{-2x}$ and now have $5^{x(k+2)}$ since the bottom half ended up being $1$ and the top ends up $5^{x(k+2)} - 0$. But I just don't know what to make of it from here. Could $k$ be any number except $2$?
One more hint:
$$\frac{5^{kx}-1}{5^{-2x} + 1} = \frac{5^{kx}}{5^{-2x} + 1} - \frac{1}{5^{-2x} + 1} = \frac{1}{5^{-(2+k)x} + 5^{-kx}} - \frac{1}{5^{-2x} + 1} $$