Answer
The solution is
$$\lim_{x\to 1}\frac{\ln x}{4x-x^3+3}=\frac{1}{2}.$$
Work Step by Step
We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{x\to 1}\frac{\ln x}{4x-x^3+3}=\left[\frac{0}{0}\right][\text{LH}]=\lim_{x\to 1}\frac{(\ln x)'}{(4x-x^2-3)'}=\lim_{x\to1}\frac{\frac{1}{x}}{4-2x}=\lim_{x\to1}\frac{1}{4x-2x^2}=\left[\frac{1}{4\cdot1-2\cdot1^2}\right]=\frac{1}{2}.$$
