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

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

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limit as x approaches 0+of tan(x)(ln(x))

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

0

Solution steps

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

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{t a n ( x )}})

Apply L'Hopital's Rule

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

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

Simplify

\frac{\frac{1}{x}}{- csc ^{2} ( x )} : - \frac{1}{x csc ^{2} ( x )}

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

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

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

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

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

x \to 0 + lim (1) = 1

x \to 0 + lim (x csc^{2} (x)) = \infty

= - \frac{1}{\infty}

Apply Infinity Property:

\frac{c}{\infty} = 0

= 0

x \to 1 + lim (x^{\frac{6}{( x - 1 )}})

x \to 0 + lim (\frac{x}{sin ( 8 x )})

x \to - 3 lim (5 x + 9)

x \to 0 lim (\frac{sin ^{1} ( x )}{x})

x \to 1 lim (∣ x - 3 ∣)

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