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?

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 $$l = \lim_{y\to0}\left(\dfrac{\ln(1+y)}{3^y-5^{-y}}\right)$$ $$ = \lim_{y\to0}\left(\dfrac{y-\dfrac{y^2}{2}+\cdots}{(1+y\ln3+\cdots)-(1-y\ln5+\cdots)}\right)$$ $$=\dfrac{1}{\ln3+\ln5}=\dfrac{1}{\ln15}$$

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:

$g(x)=\dfrac{15^y-1}{5^y}=$

$5^{-y}(15^y-1)=$

$5^{-y}(y\log 15)\dfrac{e^{y\log 15}-1}{y\log 15}$;

$\dfrac{f(x)}{g(x)}=$

$[\dfrac{\log (y+1)-\log 1}{y}]\cdot$

$[\dfrac{5^y}{\log 15}]$