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

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

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

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

0

Solution steps

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

Rewrite for L'Hopital

= x \to 0 + lim (\frac{ln ( 2 x )}{\frac{1}{s i n ( x )}})

Apply L'Hopital's Rule

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

= 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 )}

= 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)

= - 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

= - \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

= - \frac{1}{\infty}

Apply Infinity Property:

\frac{c}{\infty} = 0

= 0

\frac{d ^{3}}{d x ^{3}} (x^{4} - 4 x^{3})

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

\frac{\partial}{\partial y} (cos (y) sin (x))

(- 3 t^{2}) (t - 2)

25 x^{3} - 6 x^{2} + 7 x - 88

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