límite cuando x tiende a 0+de (ln(x))/(csc(x))

Solución

x \to 0 + lim (\frac{ln ( x )}{csc ( x )})

0

Pasos de solución

x \to 0 + lim (\frac{ln ( x )}{csc ( x )})

Utilizar la identidad trigonométrica básica:

\frac{1}{csc ( x )} = sin (x)

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

Reescribir para ajustar por L’Hopital

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

Aplicar la regla de L'Hopital

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

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

Simplificar

\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

con excepción de forma indeterminada

= - \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}

Aplicar las propiedades para limites infinitos/en el infinito:

\frac{c}{\infty} = 0

= 0

f (x) = \frac{1}{( 1 - x ) ^{2}}

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

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

y^{^{''}} - y = \frac{e ^{x}}{e ^{x} + 1}

\frac{d}{d x} (ln (\frac{x + 1}{x - 2}))