I tried to solve this limit $$\lim_{x\to 1} \left(\frac{x}{x-1}-\frac{1}{\ln x}\right)$$ and, without thinking, I thought the result was 1. But, using wolfram to verify, I noticed that the limit is $1/2$.
How can I solve it without Hopital/series/integration, just with known limits (link in comments) /squeeze/basic therem?
I don't have a clue about what I can do, because known limits can't be used ($\ln x = x-1$) without considering the "error" that we don't exactly know!
On
Note that $$\ln(x)=\int_1^x \frac{dt}{t}$$
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 $$0 < \frac{(3-x)(x-1)}{2} < \ln(x) < \frac{(1+\frac{1}{x})(x-1)}{2}$$ for all $x\in(1,3)$. Therefore on the same interval
$$ \frac{2-x}{3-x}<\frac{x}{x-1}-\frac{1}{\ln(x)}< \frac{x}{x+1} $$
and $$\lim_{x\to 1^+}\left(\frac{x}{x-1}-\frac{1}{\ln(x)}\right)=\frac{1}{2}.$$
For $x\in(0,1)$ all inequalities change direction and therefore also
$$\lim_{x\to 1^-}\left(\frac{x}{x-1}-\frac{1}{\ln(x)}\right)=\frac{1}{2}.$$
Hence we will get
$$\lim_{x\to 1}\left(\frac{x}{x-1}-\frac{1}{\ln(x)}\right)=\frac{1}{2}.$$
Set $x=e^{y}$, then $y\to 0$ implies $x\to1$. So $$ \frac{x}{x-1}-\frac1{\ln(x)}=\frac1{1-e^{y}}-\frac1y $$ Using the power series of the exponential, $e^{-y}=1-y+\frac12y^2+O(y^3)$, one can transform to $$ =\frac1{y\bigl(1-\frac12y+O(y^2)\bigr)}-\frac1y=\frac{(1+\frac12y+O(y^2))-1}y=\frac12+O(y) $$