What is the limit as x approaches 0+of xln(tan(9x)) ?

The limit as x approaches 0+of xln(tan(9x)) 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(9x))

x \to 0 + lim (x ln (tan (9 x)))

0

Solution steps

x \to 0 + lim (x ln (tan (9 x)))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( tan ( 9 x ) )}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to 0 + lim (\frac{9 sec ^{2} ( 9 x ) cot ( 9 x )}{- \frac{1}{x ^{2}}})

= x \to 0 + lim (\frac{9 sec ^{2} ( 9 x ) cot ( 9 x )}{- \frac{1}{x ^{2}}})

Simplify

= x \to 0 + lim (- 9 x^{2} sec^{2} (9 x) cot (9 x))

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

= - 9 \cdot x \to 0 + lim (x^{2} sec^{2} (9 x) cot (9 x))

Express with sin, cos

= - 9 \cdot x \to 0 + lim (\frac{x ^{2}}{cos ( 9 x ) sin ( 9 x )})

\frac{x ^{2}}{cos ( 9 x ) sin ( 9 x )} = \frac{x ^{2}}{sin ( 9 x )} \cdot \frac{1}{cos ( 9 x )}

= - 9 \cdot x \to 0 + lim (\frac{x ^{2}}{sin ( 9 x )} \cdot \frac{1}{cos ( 9 x )})

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

With the exception of indeterminate form

= - 9 \cdot x \to 0 + lim (\frac{x ^{2}}{sin ( 9 x )}) \cdot x \to 0 + lim (\frac{1}{cos ( 9 x )})

x \to 0 + lim (\frac{x ^{2}}{sin ( 9 x )}) = 0

x \to 0 + lim (\frac{1}{cos ( 9 x )}) = 1

= - 9 \cdot 0 \cdot 1

Simplify

= 0

d y - (y - 2)^{2} d x = 0

\int_{- 2 π}^{π} cos^{2} (x) d x

x \to \infty lim ((1 + (\frac{2}{x}))^{3 x})

x \to 4 lim (\frac{- 1}{x ^{2} - 16})

\frac{( x - 1 ) ^{2}}{x ^{2} + 1}

What is the limit as x approaches 0+of xln(tan(9x)) ?
The limit as x approaches 0+of xln(tan(9x)) is 0