What is the limit as x approaches 0+of ln(1-cos(x))-ln(sin(x)) ?

The limit as x approaches 0+of ln(1-cos(x))-ln(sin(x)) is -infinity

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 0+of ln(1-cos(x))-ln(sin(x))

x \to 0 + lim (ln (1 - cos (x)) - ln (sin (x)))

- \infty

Solution steps

x \to 0 + lim (ln (1 - cos (x)) - ln (sin (x)))

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (1 - cos (x)) - ln (sin (x)) = ln (\frac{1 - cos ( x )}{sin ( x )})

= x \to 0 + lim (ln (\frac{1 - cos ( x )}{sin ( x )}))

Apply the Limit Chain Rule:

- \infty

= - \infty

\frac{d}{d x} (18 x)

\frac{d y}{d t} + 2 y = 1, y (0) = \frac{5}{2}

\frac{\partial}{\partial z} (x y - 3 x z + yz)

\frac{d ^{3}}{d x ^{3}} (x^{4} - 3 x^{2} + 6 x - 5)

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

What is the limit as x approaches 0+of ln(1-cos(x))-ln(sin(x)) ?
The limit as x approaches 0+of ln(1-cos(x))-ln(sin(x)) is -infinity