I want to calculate $\displaystyle\lim_{x\to a}\left(\frac{f(x)}{f(a)}\right)^{\frac{1}{\ln{x}-\ln{a}}}$, where $f$ is a differentiable function at $x=a$ with $f(a)>0$.
I found this exercises in some calculus notes in the section of derivatives, so I guess the objective of the exercise is to write the expression as the limit definition of derivatives. My guess is there is a typo, because if the fraction of the exponent is $\frac{1}{x-a}$ instead it's direct writing the expression as $\exp(\ln(g(x))$ with $g$ the function in the limit.
Even if there is a typo I would like to know how to calculate this limit. My tries has been writing that expression as $\exp(\ln(g(x))$, aply continuity of exponential and use logarithm properties, but I haven't got anything useful.
Also, I would appreciate answers without L'hopital as this exercise is at the start of the section of derivatives of the notes.
Because of the exponent, my immediate instinct is to take the logarithm. If $$ L :=\lim_{x\to a}\left(\frac{f(x)}{f(a)}\right)^{\frac{1}{\ln{x}-\ln{a}}} $$ then by continuity and logarithm properties, $$ \ln(L)= \lim_{x \to a} \frac{\ln(f(x)) - \ln(f(a))}{\ln(x)-\ln(a)} $$ One may note that $$ \frac{\ln(f(x)) - \ln(f(a))}{\ln(x)-\ln(a)} = \frac{\ln(f(x)) - \ln(f(a))}{x-a} \cdot \frac{x-a}{\ln(x)-\ln(a)} $$ and use the definition of derivative and solve for $L$ to conclude.