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

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

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limit as x approaches 0+of tan(x^2)ln(x)

x \to 0 + lim (tan (x^{2}) ln (x))

0

Solution steps

x \to 0 + lim (tan (x^{2}) ln (x))

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{t a n ( x ^{2} )}})

Apply L'Hopital's Rule

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

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

Simplify

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

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

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

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

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

x \to 0 + lim (1) = 1

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

= - \frac{1}{2} \cdot \frac{1}{\infty}

Simplify

- \frac{1}{2} \cdot \frac{1}{\infty} : 0

= 0

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