What is the limit as x approaches 0+of xln(tan(4x)) ?

The limit as x approaches 0+of xln(tan(4x)) is 0

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limit as x approaches 0+of xln(tan(4x))

x \to 0 + lim (x ln (tan (4 x)))

0

Solution steps

x \to 0 + lim (x ln (tan (4 x)))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( tan ( 4 x ) )}{\frac{1}{x}})

Apply L'Hopital's Rule

= x \to 0 + lim (\frac{4 sec ^{2} ( 4 x ) cot ( 4 x )}{- \frac{1}{x ^{2}}})

Simplify

= x \to 0 + lim (- 4 x^{2} sec^{2} (4 x) cot (4 x))

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

= - 4 \cdot x \to 0 + lim (x^{2} sec^{2} (4 x) cot (4 x))

Express with sin, cos

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

\frac{x ^{2}}{cos ( 4 x ) sin ( 4 x )} = \frac{x ^{2}}{sin ( 4 x )} \cdot \frac{1}{cos ( 4 x )}

= - 4 \cdot x \to 0 + lim (\frac{x ^{2}}{sin ( 4 x )} \cdot \frac{1}{cos ( 4 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

= - 4 \cdot x \to 0 + lim (\frac{x ^{2}}{sin ( 4 x )}) \cdot x \to 0 + lim (\frac{1}{cos ( 4 x )})

x \to 0 + lim (\frac{x ^{2}}{sin ( 4 x )}) = 0

x \to 0 + lim (\frac{1}{cos ( 4 x )}) = 1

= - 4 \cdot 0 \cdot 1

Simplify

= 0

What is the limit as x approaches 0+of xln(tan(4x)) ?
The limit as x approaches 0+of xln(tan(4x)) is 0