What is the limit as n approaches 0 of 1/n sin(1/n) ?

The limit as n approaches 0 of 1/n sin(1/n) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as n approaches 0 of 1/n sin(1/n)

n \to 0 lim (\frac{1}{n} sin (\frac{1}{n}))

diverges

Solution steps

n \to 0 lim (\frac{1}{n} sin (\frac{1}{n}))

Simplify

\frac{1}{n} sin (\frac{1}{n}) : \frac{sin ( \frac{1}{n} )}{n}

= n \to 0 lim (\frac{sin ( \frac{1}{n} )}{n})

Apply Limit Divergence Criterion:

diverges

= diverges

x \to 0 + lim (- x)

\frac{d}{d x} (\frac{3}{( 4 x - 5 )})

\int 2 x (x^{2} - 1)^{102} d x

\int \frac{2}{1 - x ^{2}} d x

\int - sin^{2} (t) d t

What is the limit as n approaches 0 of 1/n sin(1/n) ?
The limit as n approaches 0 of 1/n sin(1/n) is diverges