MCQ
$\mathop {\lim }\limits_{x \to a} \frac{{({x^{ - 1}} - {a^{ - 1}})}}{{x - a}} = $
- A$1/a$
- B$\frac{{ - 1}}{a}$
- C$\frac{1}{{{a^2}}}$
- ✓$\frac{{ - 1}}{{{a^2}}}$
✓
Answer
Correct option: D.
$\frac{{ - 1}}{{{a^2}}}$
d
(d) $\mathop {\lim }\limits_{x \to a} \,\frac{{(1/x) - (1/a)}}{{x - a}} = \mathop {\lim }\limits_{x \to a} \,\frac{{a - x}}{{ax\,(x - a)}} $
(d) $\mathop {\lim }\limits_{x \to a} \,\frac{{(1/x) - (1/a)}}{{x - a}} = \mathop {\lim }\limits_{x \to a} \,\frac{{a - x}}{{ax\,(x - a)}} $
$= \mathop {\lim }\limits_{x \to a} \,\frac{{ - 1}}{{ax}} = \frac{{ - 1}}{{{a^2}}}$.
Or $\mathop {\lim }\limits_{x \to a} \,\,\frac{-1/{{x}^{2}}-0}{1-0}$
(By Appling $L-$ Hospital’s rule)
$-\frac{1}{{{a}^{2}}}$
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