Although I sort of have an intuition that this might be $\ln(2)$, I can't really get there.
I tried to rewrite it as :
$$\lim\limits_{x\to \infty} x \ln5 + \lim\limits_{x\to \infty} \ln(2) - \lim\limits_{x\to \infty}\ln(5^x - 1) = \infty + \ln(2) - \infty$$
Which, as far as I know, is one of the indeterminate forms$(\infty - \infty)$. How do I go about this?
On
Hint: $$ x\ln5-\ln(5^x-1)=\ln\left(\frac{5^x}{5^x-1}\right) $$ and the logarithm is continuous at $1$.
On
Hint: $x\ln5=\ln5^ x$
Thus $\lim_{x\to \infty}( \ln2+x\ln5-\ln{(5^x-1)})=\lim_{x\to \infty} (ln2+\ ln5^x-\ln{(5^x-1)})=\lim_{x\to \infty} (\ln2+\ln {\frac{5^x}{5^x-1}})=\lim_{x\to \infty} \ln2+\ln (lim_{x\to \infty}{ (\frac {5^x}{5^x-1})})=\ln2+\ln1=\ln2$
On
Assuming that you want to go further $$y=x\ln(5) + \ln(2) - \ln(5^x - 1)=\ln(2)+\ln(5^x )-\ln(5^x - 1)=\ln(2)-\ln\left(1-\frac 1 {5^x} \right)$$ Now, use equivalents to get $$y\sim \ln(2)+\frac 1 {5^x}$$ which shows the limit and how it is approached.
Use it even for $x=5$ (far away from $\infty$); the exact formula would give $$y=\log (2)+5 \ln (5)-\ln (3124)\approx 0.69346723$$ while the approximation would give $$y\sim \log (2)+\frac 1 {3125}\approx 0.69346718$$
\begin{align}\ln 2 + \lim_{x \to \infty} (x\ln 5 - \ln(5^x-1))&= \ln 2 + \lim_{x \to \infty} \ln \left(\frac{5^x}{5^x-1}\right)\\ &= \ln 2 + \lim_{x \to \infty}\ln\left(\frac{1}{1-5^{-x}}\right)\\&= \ln 2\end{align}