<s>[INST] Solution verification:$\lim_{x\to 2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$ 
Evaluate without L'Hospital:$$\lim_{x\to
 2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$$

My attempt:
I used: $$\lim_{f(x)\to 0}\frac{\ln(1+f(x))}{f(x)}=1\;\&\;\lim_{f(x)\to 0}\frac{a^{f(x)}-1}{f(x)}=\ln a$$
$$
\begin{split}
L &= \lim_{x\to 2} \frac{\ln(x-1)}{3^{x-2}-5^{-x+2}} \\
  &= \lim_{x\to 2} \frac{\dfrac{\ln(1+(x-2))}{x-2}\cdot(x-2)}
                        {(x-2)\cdot\dfrac{3^{x-2}-1+1-5^{-x+2}}{x-2}} \\
  &= \lim_{x\to 2} \frac{\dfrac{\ln(1+(x-2))}{x-2}}
                        {\dfrac{3^{x-2}-1}{x-2}+\dfrac{5^{2-x}-1}{2-x}} \\
  &=\frac{1}{\ln3+\ln5} \\
  &=\frac{1}{\ln(15)}
\end{split}
$$
Is this correct?
 [/INST] This is fine. Here is an alternative approach using taylor series:
First substitute $x-2=y$ to simplify it. Let the required limit be $l$. Then
$$l = \lim_{y\to0}\left(\dfrac{\ln(1+y)}{3^y-5^{-y}}\right)$$
$$ = \lim_{y\to0}\left(\dfrac{y-\dfrac{y^2}{2}+\cdots}{(1+y\ln3+\cdots)-(1-y\ln5+\cdots)}\right)$$
$$=\dfrac{1}{\ln3+\ln5}=\dfrac{1}{\ln15}$$
</s>