mathop {lim }limits_{x to a} frac{{({x^{ - 1} | vedclass

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Question

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o a} \frac{x^{-1} - a^{-1}}{x - a}$.Step $1$: Simplify the expression inside the limit

Vedclass · Accepted Answer

$-1/a^2$

Vedclass · Answer

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o a} \frac{x^{-1} - a^{-1}}{x - a}$.Step $1$: Simplify the expression inside the limit.$\frac{x^{-1} - a^{-1}}{x - a} = \frac{\frac{1}{x} - \frac{1}{a}}{x - a} = \frac{\frac{a - x}{ax}}{x - a}$.Step $2$: Simplify the fraction.$\frac{-(x - a)}{ax(x - a)} = -\frac{1}{ax}$ (for $x 
eq a$).Step $3$: Apply the limit as $x 	o a$.$\mathop {\lim }\limits_{x 	o a} (-\frac{1}{ax}) = -\frac{1}{a(a)} = -\frac{1}{a^2}$.Alternatively,using $L'	ext{Hospital's rule}$:$\mathop {\lim }\limits_{x 	o a} \frac{\frac{d}{dx}(x^{-1} - a^{-1})}{\frac{d}{dx}(x - a)} = \mathop {\lim }\limits_{x 	o a} \frac{-x^{-2}}{1} = -\frac{1}{a^2}$.

$\mathop {\lim }\limits_{x \to a} \frac{{({x^{ - 1}} - {a^{ - 1}})}}{{x - a}} = $

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