Find the limit: $\lim \limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right)$ Find:
$$\lim\limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right)
$$
Without using L'Hospital or Taylor approximations
Thanks in advance
 A: Note that $$\log(x)=\int_1^x \frac{dt}{t}$$
and that for $1 < t < x$ $$2-t < \frac{1}{t} < \frac{\frac{1}{x}-1}{x-1}(t-1) +1.$$
Integration in $t$ from $1$ to $x$ results in $$0 < \frac{(3-x)(x-1)}{2} < \log(x) < \frac{(1+\frac{1}{x})(x-1)}{2}$$ for all $x\in(1,3)$.  Therefore on the same interval
$$ \frac{2-x}{3-x}<\frac{x}{x-1}-\frac{1}{\log(x)}< \frac{x}{x+1} $$
and $$\lim_{x\downarrow 1}\left(\frac{x}{x-1}-\frac{1}{\log(x)}\right)=\frac{1}{2}.$$
For $x\in(0,1)$ all inequalities change direction and therefore also
$$\lim_{x\uparrow 1}\left(\frac{x}{x-1}-\frac{1}{\log(x)}\right)=\frac{1}{2}.$$
A: Let's write
$$\lim_{x\rightarrow1}\left({x\over x-1}-{1\over\ln x}\right)=1+\lim_{x\rightarrow1}\left({1\over x-1}-{1\over\ln x}\right)=1+L$$
and see if we can compute $L$.  But since $x\rightarrow1$ is equivalent to $x^2\rightarrow1$, we can rewrite $L$ as
$$L=\lim_{x\rightarrow1}\left({1\over x^2-1}-{1\over\ln x^2}\right)={1\over2}\lim_{x\rightarrow1}\left({-1\over x+1}+{1\over x-1}-{1\over\ln x}\right)={-1\over4}+{1\over2}L$$
So if we assume the limit exists, then we easily find $L=-1/2$, hence the desired limit is $1-1/2=1/2$.  Please note, though, this doesn't prove that the limit exists, just what its value is if it does.
