How do I solve with the logarithm in the exponent?
$$ \lim_{{x\to1}} \left(\log_{5}5x\right)^{\log_{x}5}$$
I'm not adept enough to understand how to deal with it.
My approach
$$\lim_{x\to1}\left(\log_{5}5x\right)^{\log_{x}5}=\lim_{x\to1}e^{\log_{x}5\log\left(\log_{5}5x\right)}=e^{\lim_{x\to1}\log_{x}5\log\left(\log_{5}5x\right)}$$
- LHL
$$\lim_{x\to1^-}\log_{x}5\log\left(\log_{5}5x\right) =e^1=e$$
- RHL
$$\lim_{x\to1^+}\log_{x}5\log\left(\log_{5}5x\right) =e^1=e$$
or
$$e^{\lim_{x\to1}\log_{x}5\left(\log_{5}5x-1\right)} =e^{\lim_{x\to1}\log_{x}5\left(\log_{5}5 +\log_5 x-1\right)}=e^{\lim_{x\to1}\log_{x}5\log_5 x}=e^1=e$$
Best Answer
\begin{align*} \lim_{{x\to1}} \left(\log_{5}5x\right)^{\log_{x}5} &= \lim_{x \to 1} \left( \log_5 5+\log_5 x \right)^{\log_x 5}\\ &= \lim_{x \to 1} \left( 1+ \frac{1}{\log_x 5} \right)^{\log_x 5}\\ &= \lim_{t \to \infty} \left( 1+ \frac{1}{t}\right)^{t}\quad,\quad t = \log_{x}5 \to \infty\;\text{ as }\;x\to1\\ &= e \end{align*}