límite cuando x tiende a 0+de ln(x)sin^2(x)

Solución

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

0

Pasos de solución

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

Reescribir para ajustar por L’Hopital

= x \to 0 + lim (\frac{sin ^{2} ( x )}{\frac{1}{l n ( x )}})

Aplicar la regla de L'Hopital

: x \to 0 + lim (\frac{sin ( 2 x )}{- \frac{1}{x l n ^{2} ( x )}})

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

Simplificar

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

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

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

Aplicar la siguiente propiedad algebraica

: a \cdot b = \frac{a}{\frac{1}{b}}, b \neq = 0

ln^{2} (x) sin (2 x) x = \frac{ln ^{2} ( x ) x}{\frac{1}{s i n ( 2 x )}}

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

con excepción de forma indeterminada

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

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

x \to 0 + lim (\frac{1}{sin ( 2 x )}) = \infty

= - \frac{0}{\infty}

Aplicar las propiedades para limites infinitos/en el infinito:

\frac{0}{\infty} = 0

= 0

x \to 0 lim (3 x^{2})

x \to \infty lim (\frac{2 x ^{3} - 5 x ^{2} + 4 x - 345}{6 x ^{3} + 2 x})

x \to 0 lim (4 + e^{\frac{1}{x}})

x \to 90 o lim (sin (x))

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