Limit of $\lim\limits_{x \to -3} \frac {4^\frac{x+3}{5}-1}{x+3}$, without L'Hopital's rule. I tried substituting $x+3$ to see if I could simplify in any way, but couldn't think of anything. Also tried using $\ln$ and $\exp$, but in the end just got to $\ln(0)$. Can someone give me a tip?
 A: HINT
Are you acquainted to the derivative definition?
\begin{align*}
\lim_{x\to-3}\frac{4^{\frac{x+3}{5}} - 1}{x + 3} = \lim_{x\to-3}\frac{4^{\frac{x+3}{5}} - 4^{\frac{-3 + 3}{5}}}{x - (-3)}
\end{align*}
Can you take it from here?
A: If you prefer a proof without using derivatives: Observe that our limit is also equal to
$$\lim\limits_{x\to-3}\frac {4^{(x+3)/5}-1}{x+3}=\frac 15\lim\limits_{x\to0}\frac {4^x-1}x$$
Where the substitution $x\mapsto\tfrac 15(x+3)$ was made. The resulting limit can be tackled by using another substitution
$$n=4^x-1\qquad\implies\qquad x=\log_4(n+1)$$
Therefore, when $x$ tends towards zero, $n$ also tends towards zero. Substituting gives
$$\begin{align*}\lim\limits_{x\to0}\frac {4^x-1}x & =\lim\limits_{n\to0}\frac {n}{\log_4(n+1)}\\ & =\lim\limits_{n\to0}\frac 1{\log_4(n+1)^{1/n}}\end{align*}$$
Now use the limit definition of $e$
$$e=\lim\limits_{n\to0}(n+1)^{1/n}$$
To get
$$\lim\limits_{x\to0}\frac {4^x-1}x=\frac 1{\log_4 e}=\log 4$$
Note that $\log(\cdot)$ is the natural logarithm. Our original limit is $1/5$ of that, so we get
$$\lim\limits_{x\to-3}\frac {4^{(x+3)/5}-1}{x+3}\color{blue}{=\frac 15\log 4}$$
