What is the limit as x approaches 0+of xln(tan(2x)) ?

The limit as x approaches 0+of xln(tan(2x)) is 0

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 xln(tan(2x))

x \to 0 + lim (x ln (tan (2 x)))

0

Solution steps

x \to 0 + lim (x ln (tan (2 x)))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( tan ( 2 x ) )}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{2 sec ^{2} ( 2 x ) cot ( 2 x )}{- \frac{1}{x ^{2}}})

= x \to 0 + lim (\frac{2 sec ^{2} ( 2 x ) cot ( 2 x )}{- \frac{1}{x ^{2}}})

Simplify

= x \to 0 + lim (- 2 x^{2} sec^{2} (2 x) cot (2 x))

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

= - 2 \cdot x \to 0 + lim (x^{2} sec^{2} (2 x) cot (2 x))

Express with sin, cos

= - 2 \cdot x \to 0 + lim (\frac{x ^{2}}{cos ( 2 x ) sin ( 2 x )})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{2 x}{2 cos ( 4 x )})

= - 2 \cdot x \to 0 + lim (\frac{2 x}{2 cos ( 4 x )})

Plug in the value

x = 0

= - 2 \cdot \frac{2 \cdot 0}{2 cos ( 4 \cdot 0 )}

Simplify

- 2 \cdot \frac{2 \cdot 0}{2 cos ( 4 \cdot 0 )} : 0

= 0

\int x (2 sec^{2} (x) + tan^{2} (x)) d x

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4

x \to \frac{π}{4} lim (\frac{sin ( x )}{x})

\frac{d y}{d t} = 2 y

\frac{d}{d y} (4 - y^{2})

What is the limit as x approaches 0+of xln(tan(2x)) ?
The limit as x approaches 0+of xln(tan(2x)) is 0