What is the limit as x approaches pi/2+of-ln|cos(x)| ?

The limit as x approaches pi/2+of-ln|cos(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 pi/2+of-ln|cos(x)|

x \to \frac{π}{2} + lim (- ln ∣ cos (x) ∣)

\infty

Solution steps

x \to \frac{π}{2} + lim (- ln ∣ cos (x) ∣)

cos (x)

is negative when

x \to \frac{π}{2} +

. Therefore

∣ cos (x) ∣ = - cos (x)

= x \to \frac{π}{2} + lim (- ln (- cos (x)))

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= - x \to \frac{π}{2} + lim (ln (- cos (x)))

Apply the Limit Chain Rule:

- \infty

= - (- \infty)

Simplify

= \infty

x \to 0 lim (\frac{x ^{3} + x ^{2}}{x})

x \to 1 lim (x^{n})

x \to - 4 lim (\frac{x + 3}{x + 4})

x \to 0 lim (\frac{1 2 ^{4 x} - 1}{2 x})

x \to 5 lim (x^{2} + 11)

What is the limit as x approaches pi/2+of-ln|cos(x)| ?
The limit as x approaches pi/2+of-ln|cos(x)| is infinity