The value of lim _{x rightarrow 0} frac{1}{x^ | vedclass

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(B) Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule.Using the Leibniz integral rule,the derivative of the numerator is $\fr

Vedclass · Accepted Answer

$\frac{1}{12}$

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(B) Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's rule.Using the Leibniz integral rule,the derivative of the numerator is $\frac{d}{dx} \int_0^x \frac{t \ln (1+t)}{t^4+4} dt = \frac{x \ln (1+x)}{x^4+4}$.The derivative of the denominator $x^3$ is $3x^2$.Thus,the limit becomes $\lim _{x \rightarrow 0} \frac{x \ln (1+x)}{3x^2(x^4+4)}$.Simplifying,we get $\lim _{x \rightarrow 0} \frac{1}{3} \cdot \frac{\ln (1+x)}{x} \cdot \frac{1}{x^4+4}$.Using the standard limit $\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x} = 1$,we have $\frac{1}{3} \cdot 1 \cdot \frac{1}{0^4+4} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$.

The value of $\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^x \frac{t \ln (1+t)}{t^4+4} dt$ is

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