What is the limit as x approaches 0+of 4sin(x)ln(x) ?

The limit as x approaches 0+of 4sin(x)ln(x) is 0

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limit as x approaches 0+of 4sin(x)ln(x)

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

0

Solution steps

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

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

= 4 \cdot x \to 0 + lim (sin (x) ln (x))

Rewrite for L'Hopital

= 4 \cdot x \to 0 + lim (\frac{ln ( x )}{\frac{1}{s i n ( x )}})

Apply L'Hopital's Rule

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

= 4 \cdot x \to 0 + lim (\frac{\frac{1}{x}}{- cot ( x ) csc ( x )})

Simplify

\frac{\frac{1}{x}}{- cot ( x ) csc ( x )} : - \frac{1}{x cot ( x ) csc ( x )}

= 4 \cdot x \to 0 + lim (- \frac{1}{x cot ( x ) csc ( x )})

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

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

= 4 (- \frac{lim _{x \to 0 +} ( 1 )}{lim _{x \to 0 +} ( x cot ( x ) csc ( x ) )})

x \to 0 + lim (1) = 1

x \to 0 + lim (x cot (x) csc (x)) = \infty

= 4 (- \frac{1}{\infty})

Simplify

4 (- \frac{1}{\infty}) : 0

= 0

\frac{1}{1 + x ^{2}}

x \to 3 lim (\frac{- 4}{2 x - 5})

\frac{d}{d x} (x + 6 x^{\frac{2}{3}})

x y^{^{'}} = x y + 2 y

\frac{d ^{2}}{d x ^{2}} (x + 2)

What is the limit as x approaches 0+of 4sin(x)ln(x) ?
The limit as x approaches 0+of 4sin(x)ln(x) is 0