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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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}} = $

Similar Questions

If $\lim _{x}$ ${\rightarrow 1} \frac{(5 x+1)^{1 / 3}-(x+5)^{1 / 3}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}=\frac{m \sqrt{5}}{n(2 n)^{2 / 3}}$,where $\operatorname{gcd}(m, n)=1$,then $8 m+12 n$ is equal to.

If $f(1) = 1$ and $f'(1) = 2$,then $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$ is

Evaluate the limit: $\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$

If $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$,then

$\lim _{x \rightarrow 0} \frac{x 2^{x}-x}{1-\cos x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module