Question

consider the function $f(x)=\frac{ln x}{sin x}$. if you are human, find the limit of $f(x)$ as $x$ approaches $0$ from the right.

Explanation:

Step1: Use L'Hopital's Rule

We have the form $\frac{-\infty}{0}$ (since $\lim_{x
ightarrow0^{+}}\ln x=-\infty$ and $\lim_{x
ightarrow0^{+}}\sin x = 0$). But we can rewrite it as $\frac{\ln x}{\sin x}=\frac{\ln x}{\frac{\sin x}{x}\cdot x}$. We know that $\lim_{x
ightarrow0}\frac{\sin x}{x} = 1$. So we consider the limit $\lim_{x
ightarrow0^{+}}\frac{\ln x}{x}$. Now, it is in the form $\frac{-\infty}{\infty}$. By L'Hopital's Rule, if $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}$ is in the form $\frac{\pm\infty}{\pm\infty}$ or $\frac{0}{0}$, then $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$. Let $f(x)=\ln x$ and $g(x)=x$. Then $f^{\prime}(x)=\frac{1}{x}$ and $g^{\prime}(x) = 1$.

Step2: Calculate the limit

$\lim_{x
ightarrow0^{+}}\frac{\ln x}{x}=\lim_{x
ightarrow0^{+}}\frac{\frac{1}{x}}{1}=\lim_{x
ightarrow0^{+}}\frac{1}{x}=-\infty$

Answer:

$-\infty$