How do I solve $\ \lim_{{x\to1}} \left(\log_{5}5x\right)^{\log_{x}5} $?

Question

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.

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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

$$e^{\lim_{x\to1}\log_{x}5\left(\log_{5}5x-1\right)} =e^{\lim_{x\to1}\log_{x}5\left(\log_{5}5 +\log_5 x-1\right)}=e^{\lim_{x\to1}\log_{x}5\log_5 x}=e^1=e$$

Answer 1

\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*}

Answer 2

As an alternative to the very fine solution given, your way to start is fine with

$$\left(\log_{5}5x\right)^{\log_{x}5}=e^{\log_{x}5\log\left(\log_{5}5x\right)}$$

but not the conclusion since $\log_{x}5\log\left(\log_{5}5x\right)$ is in the form $\infty\cdot 0$.

From here we can instead proceed as follows by

$$\log_{5}5x= \frac{\log 5x}{\log 5}, \;\;\; \log_{x}5=\frac{\log 5}{\log x} $$

then

$$\left(\log_{5}5x\right)^{\log_{x}5}=e^{\log_{x}5\log\left(\log_{5}5x\right)}=e^{\frac{\log 5}{\log x}\left(\log\log (5x)-\log\log 5\right)}\to e$$

indeed by l'Hospital (very useful in this case)

$$\lim_{x\to 1}\frac{\log\log (5x)-\log\log 5}{\log x}=\lim_{x\to 1}\frac{\frac{1}{\log 5x}}{\frac1x} \to \frac{1}{\log 5}$$