# Trying to determine $\lim_{x\to\infty }\frac {x}{\sin x}$

I am trying to determine $$\lim_{x\to \infty}\frac {x}{\sin x}$$

According to the definition of limit at infinity , I think that does not exist.

Am I right? If not please correct me.

• Yes, the limit doesn't exist Jan 17 '18 at 8:37

You can see that by using the following two sequence:

1.) $x_n:= \frac{\pi}{2}+ 2\pi\cdot n$

2.) $y_n:= \frac{3\pi}{2}+ 2\pi\cdot n$

Inserting theses series into your expression gives two series, where the first diverges to $+\infty$ and the second to $-\infty$. This implies that the expression has no limit.

A similar agument shows that the limit $lim_\infty \frac{1}{sin(x)}$ does not exists, too.

• Mark.Nice answer. Jan 17 '18 at 9:08

You are right. The sine function keeps alternating between positive and negative values while the quotient becomes arbitrarily large in absolute value.
(Aside: $\lim_{x\to\infty} x/|\sin(x)| = +\infty$.)

• What's about $$\lim_{x\to \infty }\frac {1}{\sin x}$$ ? Jan 17 '18 at 8:29
• Does not exist either. Here $\varliminf_{x\to\infty} 1/|\sin(x)|=1$ and in fact, every number $\geq 1$ is a cluster point. Jan 17 '18 at 8:35
• One thing that I was found from definition of limit at infinity that there doesn't exist any $(c,\infty )\subset \mathbb D$ for some $c\in\mathbb R$ . Where $\mathbb D$ be the domain of functions. That contradicts the existence of above limits at infinity. Jan 17 '18 at 9:29