$\lim \frac{x}{\sin x} = +\infty$ or $-\infty $ as $ x \rightarrow (n\pi) , n \neq 0$ is a wrong statement.

Question

$$\lim \frac{x}{\sin x} = +\infty \textrm{ or } -\infty $$ as $ x \rightarrow (n\pi) , n \neq 0$ is a wrong statement. My professor told me this and he told me that the correct statement to write is:

$$\lim \frac{x}{\sin x} = +\infty \textrm{ or } -\infty $$ as $ x \rightarrow (n\pi)^+ , n \neq 0$

But I do not understand how this "+" above the $(n \pi)$ yields to $ +\infty$ or $-\infty $, could anyone explain this for me please?

Answer 1

We have that for $n\neq 0$

$$\lim_{x\to n\pi^+} \frac{x}{\sin x}$$

doesn't exist since it leads to a different limit for n even or odd.

Indeed let consider for example

• $x\to \pi^+ \implies \sin x \to 0^- \implies \lim_{x\to \pi^+} \frac{x}{\sin x}=-\infty$

but for

• $x\to 2\pi^+ \implies \sin x \to 0^+\implies \lim_{x\to 2\pi^+} \frac{x}{\sin x}=+\infty$.