How do I go about finding this kind of limit? $\lim\limits_{x \to 5} (x^2 - 25)^{\frac{1}{\ln(x-5)}}$
I understand that the function isn't defined for $x \le 5$, but I'm interested in finding the Right Hand Limit (RHL) i.e., limit as $x \to5^{+}$.
How do I approach this problem?
 A: \begin{align}
\lim_{x\to 5^+}(x^2-25)^{\frac{1}{\ln(x-5)}} 
&= \lim_{x\to 5^+} \left[e^{\ln(x^2-25)}\right]^{\frac{1}{\ln(x-5)}} \\
&= \lim_{x\to 5^+} \exp\left[\frac{1}{\ln(x-5)}\ln(x^2-25)\right] \\
&= \lim_{x\to 5^+} \exp\left[\frac{1}{\ln(x-5)}\Big(\ln(x+5) + \ln(x-5)\Big)\right] \\
&= \lim_{x\to 5^+} \exp\left[\frac{\ln(x+5)}{\ln(x-5)} + 1\right] \\
&= \exp\left[\lim_{x\to 5^+} \left(\frac{\ln(x+5)}{\ln(x-5)}\right) + 1\right] \\
&= e
\end{align}
A: 
$\lim\limits_{x \to 5} (x^2 - 25)^{\frac{1}{\ln(x-5)}}$

Alternative approach:
Let $u = x-5.$
$\displaystyle \lim_{u \to 0^+} f(u) = \left[(u + 10)(u)\right]^{\left(\frac{1}{\log u}\right)}.$
Focus on the logarithm.
Let $\displaystyle g(u) = \log[f(u)] = 
\frac{\log(u + 10) + \log (u)}{\log (u)}.$
$$\lim_{u \to 0^+} g(u) = 
\lim_{u \to 0^+} \left[1 + \frac{\log(u + 10)}{\log (u)}
\right].\tag1 $$
As $(u \to 0^+), ~\log(u + 10) \to \log(10)~~$ and
$~~\log(u) \to -\infty.$
Therefore, as $\displaystyle (u \to 0^+), \frac{\log(u + 10)}{\log (u)} \to 0.$
Therefore, using (1) above,
$\displaystyle \lim_{u \to 0^+} g(u) = 1.$
Therefore, $\displaystyle \lim_{u \to 0^+} f(u) = e.$
A: As an alternative, by $x-5=e^{-y}\to 0$ with $y\to \infty$ we have
$$\lim\limits_{x \to 5} (x^2 - 25)^{\frac{1}{\ln(x-5)}}=\lim\limits_{y \to \infty} \left[e^{-y}(e^{-y}+10)\right] ^{-\frac{1}{y}} =\lim\limits_{y \to \infty} \frac e{(e^{-y}+10)^{\frac{1}{y}}}=e$$
