<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] Your answer a bit rephrased.
$y=x-2;$
$\dfrac{\log (1+y)}{3^y-5^{-y}}$;
Numerator : 
$f(y):=y\dfrac{\log (1+y)-\log 1}{y}$
Denominator:
$g(x)=\dfrac{15^y-1}{5^y}=$
$5^{-y}(15^y-1)=$
$5^{-y}(y\log 15)\dfrac{e^{y\log 15}-1}{y\log 15}$;
$\dfrac{f(x)}{g(x)}=$
$[\dfrac{\log (y+1)-\log 1}{y}]\cdot$
$[\dfrac{5^y}{\log 15}]$
$ \big [\dfrac{1}{\dfrac{e^{y\log15}-1}{y\log 15}}\big ].$
Take the limit $y \rightarrow 0$.
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