Answer
$a)$ $\infty-\infty$
$b)$ $\infty$, so the limit does not exist
$c)$ see picture
Work Step by Step
$a)$ $\lim\limits_{x \to 1^{+}}(\frac{3}{lnx}-\frac{2}{x-1})=\frac{3}{ln(1^{+})}-\frac{2}{1^{+}-1}=\frac{3}{0^{+}}-\frac{2}{0^{+}}=\infty-\infty$
$\infty-\infty$ is an indeterminate form.
$b)$ $\lim\limits_{x \to 1^{+}}(\frac{3}{lnx}-\frac{2}{x-1})=\frac{3}{ln(1^{+})}-\frac{2}{1^{+}-1}=\frac{3}{0^{+}}-\frac{2}{0^{+}}=\infty-\infty$
$\infty-\infty$ is an indeterminate form. We can't do l'Hopital's Rule until the indeterminate form is either $\frac{\infty}{\infty}$ or $\frac{0}{0}$, so:
$\lim\limits_{x \to 1^{+}}(\frac{3(x-1)-2lnx}{lnx(x-1)})$
$\lim\limits_{x \to 1^{+}}(\frac{3x-1-2lnx}{lnx(x-1)})$
$\lim\limits_{x \to 1^{+}}(\frac{3-2\frac{1}{x}}{lnx+(x-1)\frac{1}{x}})$
$\lim\limits_{x \to 1^{+}}(\frac{3-2\frac{1}{x}}{lnx+(x-1)\frac{1}{x}})(\frac{x}{x})$
$\lim\limits_{x \to 1^{+}}(\frac{3x-2}{xlnx+x-1})=\frac{3(1)^{+}-2}{(1^{+})ln(1^{+})+1^{+}-1}=\frac{1^{+}}{0^{+}}=\infty$, so the limit does not exist
$c)$ see picture