mathop {lim }limits_{x to 0} left[ {frac{1}{x | vedclass

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Question

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} ight]$.Combine the terms

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$\frac{1}{2}$

Vedclass · Answer

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} ight]$.Combine the terms into a single fraction:$\mathop {\lim }\limits_{x 	o 0} \left[ {\frac{x - \log (1 + x)}{{x^2}}} ight]$.Since this is a $\frac{0}{0}$ indeterminate form,we apply $L'	ext{Hospital's rule}$:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}(x - \log (1 + x))}{\frac{d}{dx}(x^2)} = \mathop {\lim }\limits_{x 	o 0} \frac{1 - \frac{1}{1+x}}{2x}$.Simplify the expression:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{1+x-1}{1+x}}{2x} = \mathop {\lim }\limits_{x 	o 0} \frac{x}{2x(1+x)} = \mathop {\lim }\limits_{x 	o 0} \frac{1}{2(1+x)}$.Substitute $x = 0$:$\frac{1}{2(1+0)} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} \right] =$

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