What is the $\lim_\limits{x \to 1+}\left(\frac{3x}{x-1}-\frac{1}{2 \ln(x)}\right)$? 
What is the limit of $$\lim_\limits{x \to 1+} \left(\frac{3x}{x-1}-\frac{1}{2 \ln(x)} \right)$$

I attemped the problem using L^Hopital's Rule.
My Work
$$\lim_\limits{x \to 1+} \left(\frac{3x}{x-1}-\frac{1}{2 \ln(x)} \right)$$
$$\lim_\limits{x \to 1+} \left(-\frac{3}{(x-1)^2}+\frac{1}{2 \ln^2(x)\cdot x} \right)$$
$$\frac{-3+1}{0}=\frac{-2}{0}$$
The answer is suppose to be $\infty$. I know what I did is probably not right.
 A: You can only use L'Hopital's rule for limits of the form $0/0$ or $\pm\infty/\pm\infty$. So, you must first bring your function to a common denominator:
$$
\frac{3x}{x-1}-\frac{1}{2\ln x}=\frac{6x\ln x-x+1}{2x\ln x-2\ln x}.
$$
Now your limit is of the form $0/0$ and you can apply L'Hopital. 
A: Alternative more general approach. Assume that $A\in\mathbb{R}$,
$$\begin{align*}
\lim_\limits{x \to 1+}\left(\frac{Ax}{x-1}-\frac{1}{2\ln(x)}\right)
&=\lim_\limits{t \to 0+}\left(\frac{A(1+t)}{t}-\frac{1}{2\ln(1+t)}\right)\\
&=\lim_\limits{t \to 0+}\left(\frac{A}{t}+A-\frac{1}{2(t-t^2/2+o(t^2))}\right)\\
&=\lim_\limits{t \to 0+}\left(\frac{A}{t}+A-\frac{1}{2t(1-t/2+o(t))}\right)\\
&=\lim_\limits{t \to 0+}\left(\frac{A}{t}+A-\frac{1+t/2+o(t)}{2t}\right)\\
&=\lim_\limits{t \to 0+}\left(\frac{A-1/2}{t}+(A-1/4)+o(1)\right)\\
&=\left\{\begin{array}{lr}
        +\infty, & \text{if $A>1/2$,}\\
        1/4, & \text{if $A=1/2$,}\\
        -\infty, & \text{if $A<1/2$.}\\
        \end{array}\right.
\end{align*}$$
A: It is easier to put $x = 1 + h$ and then $h \to 0^{+}$. We have then
\begin{align}
L &= \lim_{x \to 1^{+}}\left(\frac{3x}{x - 1} - \frac{1}{2\log x}\right)\notag\\
&= \lim_{h \to 0^{+}}\left(\frac{3 + 3h}{h} - \frac{1}{2\log (1 + h)}\right)\notag\\
&= \lim_{h \to 0^{+}}\frac{(3 + 3h)2\log(1 + h) - h}{2h\log(1 + h)}\notag\\
&= \frac{1}{2}\lim_{h \to 0^{+}}\dfrac{6\log(1 + h) + 6h\log(1 + h) - h}{h^{2}\cdot\dfrac{\log(1 + h)}{h}}\notag\\
&= \frac{1}{2}\lim_{h \to 0^{+}}\dfrac{6\log(1 + h) - 6h + 6h\log(1 + h) + 5h }{h^{2}}\notag\\
&= \frac{1}{2}\lim_{h \to 0^{+}}\left(6\cdot\frac{\log(1 + h) - h}{h^{2}} + 6\cdot\frac{\log(1 + h)}{h} + \frac{5}{h}\right)\notag\\
\end{align}
Using Taylor expansion $$\log(1 + h) = h - \frac{h^{2}}{2} + o(h^{2})$$ we can see that the first term in parentheses tends to $6(-1/2) = -3$ and second term clearly tends to $6$. But the last term tends to $\infty$ so that the whole limit tends to $\infty$.
