- #1

ToastIQ

- 11

- 0

I'm supposed to calculate the limit of this:

\(\displaystyle \lim_{{x}\to{1}}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)\)

Combining the fractions:

\(\displaystyle \frac{x}{x-1}-\frac{1}{\ln x} = \frac{x\ln x-x+1}{(x-1)\ln x} \)

The substitute

\(\displaystyle u=x-1 \implies x=1+u \)

then gives

\(\displaystyle \frac{(1+u)\ln(1+u)-1-u+1}{(1+u-1)\ln(1+u)}\)

Maclaurin expansion of \(\displaystyle \ln(1+u) \) :

\(\displaystyle \ln(1+u) = u-\frac{u^2}{2}+u^3B(u) \)

Am I on the right path or am I completely misunderstanding this problem? It looks weird to me when I try to put it all together and I haven't been able to come to the right solution (which is \(\displaystyle \frac{1}{2} \) ).