How to evaluate $\lim _{x\to 1}\left(\frac{x+2}{x-1}-\frac{3}{\ln x}\right)$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used (without L'Hôpital if is possible)? Thanks
$$\lim _{x\to 1}\left(\frac{x+2}{x-1}-\frac{3}{\ln x}\right)$$
 A: As $x \to 1$, one may use Taylor's series expansion to get
$$
\ln x= \ln (1+(x-1))=(x-1)-\frac12(x-1)^2+O((x-1)^3)
$$ or
$$
\frac{x+2}{x-1}-\frac{3}{\ln x}=\frac{x+2}{x-1}-\frac3{(x-1)-\frac12(x-1)^2+O((x-1)^3)}
$$ that is, as $x \to 1$, 
$$
\frac{x+2}{x-1}-\frac{3}{\ln x}=-\frac12+O(x-1)
$$ from which you deduce the sought limit.
A: One possibility is to first change the variable, by setting $x=t+1$, so you get
$$
\lim_{t\to0}\left(\frac{t+3}{t}-\frac{3}{\ln(1+t)}\right)=
\lim_{t\to0}\left(1+3\left(\frac{1}{t}-\frac{1}{\ln(1+t)}\right)\right)
$$
and you can concentrate on the inner brackets, which give
$$
\lim_{t\to0}\frac{\ln(1+t)-t}{t\ln(1+t)}=
\lim_{t\to0}\frac{t-t^2/2+o(t^2)-t}{t(t+o(t))}=-\frac{1}{2}
$$
So your limit is
$$
1-\frac{3}{2}=-\frac{1}{2}
$$
A: $\mathrm{I \:am \:using \:L'Hopital's\:Rule. Given, }$
$\lim _{x\to 1}\left(\frac{x+2}{x-1}-\frac{3}{ln\left(x\right)}\right)$
$=\lim _{x\to \:1}\left(\frac{\left(x+2\right)\ln \left(x\right)-3\left(x-1\right)}{\left(x-1\right)\ln \left(x\right)}\right)=\frac{0}{0}$
$\mathrm{Apply\:L'Hopital's\:Rule:}$
$=\lim _{x\to \:1}\left(\frac{\frac{2}{x}+\ln \left(x\right)-2}{-\frac{1}{x}+\ln \left(x\right)+1}\right)$
$=\lim _{x\to \:1}\left(\frac{-2x+x\ln \left(x\right)+2}{x+x\ln \left(x\right)-1}\right)=\frac{0}{0}$
$\mathrm{Apply\:L'Hopital's\:Rule:}$
$=\lim _{x\to \:1}\left(\frac{\ln \left(x\right)-1}{\ln \left(x\right)+2}\right)$
$=\frac{\ln \left(1\right)-1}{\ln \left(1\right)+2}$
$=-\frac{1}{2}$
A: Let's put $ x = t + 1$ so that $t \to 0$ as $x \to 1$. We have then
\begin{align}
L &= \lim_{x \to 1}\left(\frac{x + 2}{x - 1} - \frac{3}{\log x}\right)\notag\\
&= \lim_{t \to 0}\left(\frac{t + 3}{t} - \frac{3}{\log(1 + t)}\right)\notag\\
&= \lim_{t \to 0}1 + 3\cdot\frac{\log(1 + t) - t}{t\log(1 + t)}\notag\\
&= 1 + 3\lim_{t \to 0}\dfrac{\log(1 + t) - t}{\dfrac{\log(1 + t)}{t}\cdot t^{2}}\notag\\
&= 1 + 3\lim_{t \to 0}\frac{\log(1 + t) - t}{t^{2}}\tag{1}\\
&= 1 + 3\lim_{t \to 0}\dfrac{\dfrac{1}{1 + t} - 1}{2t}\text{ (via L'Hospital's Rule)}\notag\\
&= 1 - \frac{3}{2}\lim_{t \to 0}\frac{1}{1 + t}\notag\\
&= 1 - \frac{3}{2}\notag\\
&= -\frac{1}{2}\notag
\end{align}
After the step marked $(1)$ we can use either L'Hospital's Rule or Taylor series.
