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

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

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

x \to 0 + lim (2 sin (x) ln (x))

0

Solution steps

x \to 0 + lim (2 sin (x) ln (x))

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

= 2 \cdot x \to 0 + lim (sin (x) ln (x))

Rewrite for L'Hopital

= 2 \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 )})

= 2 \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 )}

= 2 \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)

= 2 (- 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

= 2 (- \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

= 2 (- \frac{1}{\infty})

Simplify

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

= 0

y = x ln (x)

θ \to - \frac{π}{2} - lim (sec (θ))

\frac{d}{d x} (\frac{x}{36 + x ^{2}})

\frac{\partial}{\partial x} (2 x e^{x^{2}} + y^{2})

y^{^{''}} + 6 y^{^{'}} + 18 y = 0

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