Maclaurin Limit of lnx: \frac{1}{2}

[ToastIQ]

Hello,

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 \(\displaystyle \frac{1}{2} \) ).

 

[MarkFL]

Are you required to use a Maclaurin expansion? If not, I would use L'Hôpital's Rule instead.

 

[ToastIQ]

Are you required to use a Maclaurin expansion? If not, I would use L'Hôpital's Rule instead.

Yeah I've seen an example on this problem with L'Hôpital's Rule. But we haven't learned about L'Hôpital's Rule yet, and even if we had, we're not allowed to use it in this class.