Evaluar sin L'Hospital: $$ \ lim_ {x \ to 2} \ frac {\ ln (x-1)} {3 ^ {x-2} -5 ^ {- x + 2}} $$
Mi intento:
Usé: $$\lim_{f(x)\to 0}\frac{\ln(1+f(x))}{f(x)}=1\;\&\;\lim_{f(x)\to 0}\frac{a^{f(x)}-1}{f(x)}=\ln a$ $
$$ \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{separar} $$
¿Es esto correcto?
