How can you solve $\lim _{x \to 1}\frac{\ln\left(\frac{2x-1}{x}\right)}{\root{3}\of{x}-1}$?

Question

$\lim _{x \to 1}\frac{\ln\left(\frac{2x-1}{x}\right)}{\root{3}\of{x}-1}$

I can't use L'Hopital's Rule here. I started by substituting $\root{3}\of{x}-1$ with $t$

$\lim _{t \to 0}\frac{\ln\left(\frac{2(t+1)^3-1}{(t+1)^3}\right)}{t} \to \lim _{t \to 0}\frac{\ln(2(t+1)^3-1)-3\cdot\ln{(t+1)}}{t} \to \lim _{t \to 0}\frac{\ln(2(t+1)^3-1)}{t}-3\cdot\frac{\ln{(1+t)}}{t}$

As you can see I formated it so the right part will be $-3$ when I put in $0$. I don't know how to do the left part.

Thanks!

Answer 1

$$ \lim _{t \to 0}\frac{\ln(2(t+1)^3-1)}{t}= \lim _{t \to 0}\frac{\ln(2(t^3+3t^2+3t+1)-1)}{t} $$ $$ =\lim _{t \to 0}\frac{\ln(1+2t^3+6t^2+6t)}{t} =\lim _{t \to 0}\frac{\ln(1+2t^3+6t^2+6t)}{2t^3+6t^2+6t}\cdot \frac{2t^3+6t^2+6t}{t} $$ $$ =1\cdot \lim_{t\to 0} \left(\frac{2t^3}{t}+\frac{6t^2}{t}+\frac{6t}{t}\right) =1\cdot(0+0+6)=6. $$

Answer 2

$f(t)=\log (2(t+1)^3-1)$;

$f'(0)=\lim_{t \rightarrow 0} \dfrac{f(t)-f(0)}{t}=$

$\lim_{t \rightarrow 0}\dfrac{\log (2(t+1)^3-1)-\log 1}{t}$

$=\left (\dfrac{6(t+1)^2}{2(t+1)^3-1}\right )_{t=0}=6.$