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

The limit as x approaches 0+of (((ln(x))^2))/(ln(sin(x))) is -infinity

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

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

- \infty

Solution steps

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

Apply L'Hopital's Rule

= x \to 0 + lim (\frac{\frac{2 l n ( x )}{x}}{cot ( x )})

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

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

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

= 2 \cdot x \to 0 + lim (\frac{ln ( x )}{x cot ( 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

= 2 \cdot \frac{lim _{x \to 0 +} ( ln ( x ) )}{lim _{x \to 0 +} ( x cot ( x ) )}

x \to 0 + lim (ln (x)) = - \infty

x \to 0 + lim (x cot (x)) = 1

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

Simplify

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

= - \infty

What is the limit as x approaches 0+of (((ln(x))^2))/(ln(sin(x))) ?
The limit as x approaches 0+of (((ln(x))^2))/(ln(sin(x))) is -infinity