What is the limit as x approaches 2pi of 1/(sin(x)) ?

The limit as x approaches 2pi of 1/(sin(x)) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as x approaches 2pi of 1/(sin(x))

x \to 2 π lim (\frac{1}{sin ( x )})

diverges

Solution steps

x \to 2 π lim (\frac{1}{sin ( x )})

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

x \to 2 π + lim (\frac{1}{sin ( x )}) = \infty

x \to 2 π - lim (\frac{1}{sin ( x )}) = - \infty

= diverges

What is the limit as x approaches 2pi of 1/(sin(x)) ?
The limit as x approaches 2pi of 1/(sin(x)) is diverges