## University Calculus: Early Transcendentals (3rd Edition)

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