<s>[INST] 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.
 [/INST] 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 $\infty$.
</s>