What is $\lim_{x \to 3} (3^{x-2}-3)/(x-3)(x+5)$ without l'Hôpital's rule? I'm trying to solve the limit $\lim_{x \to 3} \frac{3^{x-2}-3}{(x-3)(x+5)}$ 
but I don't know how to proceed: $\lim_{x \to 3} \frac{1}{x+5}$ $\lim_{x \to 3} \frac{3^{x-2}-3}{x-3}$ = $1\over8$ $\lim_{x \to 3} \frac{\frac{1}{9}(3^{x}-27)}{x-3}$
Any hints? Thanks in advance.
 A: Hint:
Apply the definition of derivative
$$\lim_{x\to3}\dfrac{3^{x-2}-3}{x-3}=3\lim_{x\to3}\dfrac{3^{x-3}-1}{x-3}=3(3^{x-3})'\Big|_{x=3}=3\ln3$$
A: We have that with $y=x-3 \to 0$
$$\lim_{x \to 3} \frac{3^{x-2}-3}{(x-3)(x+5)}=\lim_{y \to 0} \frac{3\cdot 3^{y}-3}{y(y+8)}=\lim_{y \to 0} \frac{3}{y+8}\frac{3^{y}-1}{y}$$
then recall that by standard limits
$$\frac{3^{y}-1}{y}=\frac{e^{y\log 3}-1}{y\log 3}\cdot \log 3 \to \log 3$$
A: Hint:
Set $h=x-3$; $h$ tends to $0$ as $x$ tends to $3$, and you can rewrite the fraction as
$$\frac1{h+8}\,\frac{3^{h+1}-3}h=\underbrace{\frac 3{h+8}}_{\substack{\downarrow\\\tfrac38}}\underbrace{\frac{3^h-1}h}_{\substack{\downarrow\\(3^x)'_{x=0}}}.$$
A: Perhaps use definition of $3^x$ ... namely $3^x = e^{x\log 3}$.  Instaed of $x \to 3$ write $y=x-3$ and do $y \to 0$.
$$
\lim_{x \to 3} \frac{3^{x-2}-3}{(x-3)(x+5)} = \lim_{y \to 0}\frac{3^{y+1}-3}{y(y+8)}
=3 \lim_{y \to 0}\frac{3^{y}-1}{y(y+8)}
\\
3^y = \exp(y\log 3) = 1 + y\log 3 + o(y)
\\
3^y-1 = y\log 3 + o(y)
\\
y(y+8) = 8y+o(y)
\\
\frac{1}{y(y+8)} = y^{-1}\frac{1}{8}+o(y^{-1})
\\
\frac{3^y-1}{y(y+8)} = \frac{\log 3}{8} + o(y)
\\
3\frac{3^y-1}{y(y+8)} = \frac{3\log 3}{8} + o(1)
\\
3 \lim_{y \to 0}\frac{3^{y}-1}{y(y+8)} = \frac{3\log 3}{8}
$$
