<s>[INST] 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.
 [/INST] 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
$$\frac{3^{y}-1}{y}=\frac{e^{y\log 3}-1}{y\log 3}\cdot \log 3 \to \log 3$$
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