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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.

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    $\begingroup$ Yes, the limit doesn't exist $\endgroup$ Jan 17 '18 at 8:37
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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.

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  • $\begingroup$ Mark.Nice answer. $\endgroup$ Jan 17 '18 at 9:08
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You are right. The sine function keeps alternating between positive and negative values while the quotient becomes arbitrarily large in absolute value.
(Aside: $\lim_{x\to\infty} x/|\sin(x)| = +\infty$.)

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    $\begingroup$ What's about $$\lim_{x\to \infty }\frac {1}{\sin x} $$ ? $\endgroup$ Jan 17 '18 at 8:29
  • $\begingroup$ Does not exist either. Here $\varliminf_{x\to\infty} 1/|\sin(x)|=1$ and in fact, every number $\geq 1$ is a cluster point. $\endgroup$
    – max_zorn
    Jan 17 '18 at 8:35
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    $\begingroup$ 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. $\endgroup$ Jan 17 '18 at 9:29

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