The value of mathop {lim }limits_{x to 0} fra | vedclass

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Question

(B) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\log (1 + {x^3})}}{{{{\sin }^3}x}}$.Since the form is $\frac{0}{0}$ as $x

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\log (1 + {x^3})}}{{{{\sin }^3}x}}$.Since the form is $\frac{0}{0}$ as $x 	o 0$,we can use the standard limit $\mathop {\lim }\limits_{x 	o 0} \frac{{\log (1 + u)}}{u} = 1$ where $u = x^3$.$\mathop {\lim }\limits_{x 	o 0} \frac{{\log (1 + {x^3})}}{{{{\sin }^3}x}} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{{\log (1 + {x^3})}}{{{x^3}}} 	imes \frac{{{x^3}}}{{{{\sin }^3}x}} ight)$$= \mathop {\lim }\limits_{x 	o 0} \left( \frac{{\log (1 + {x^3})}}{{{x^3}}} ight) 	imes \mathop {\lim }\limits_{x 	o 0} \left( \frac{x}{{\sin x}} ight)^3$$= 1 	imes (1)^3 = 1$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\log (1 + {x^3})}}{{{{\sin }^3}x}} = $

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