Answer
$$\lim_{x\to1}g(x)=0$$
Work Step by Step
$$\lim_{x\to1}\frac{3x^2+1}{g(x)}=\infty$$
$$\frac{\lim_{x\to1}(3x^2+1)}{\lim_{x\to1}g(x)}=\infty$$
$$\frac{3\times1^2+1}{\lim_{x\to1}g(x)}=\infty$$
$$\frac{4}{\lim_{x\to1}g(x)}=\infty$$
For $\frac{4}{\lim_{x\to1}g(x)}$ to reach infinity, $\lim_{x\to1}g(x)$ needs to be equal to $0$.
$$\lim_{x\to1}g(x)=0$$
