A) \[{{x}_{n+1}}=\sqrt{2+{{x}_{n}}},\ n\ge 1,\ \]
B) \[\frac{-1}{a}\]
C) \[\frac{1}{{{a}^{2}}}\]
D) \[\frac{-1}{{{a}^{2}}}\]
Correct Answer: D
Solution :
\[\underset{x\to a}{\mathop{\lim }}\,\,\frac{(1/x)-(1/a)}{x-a}=\underset{x\to a}{\mathop{\lim }}\,\,\frac{a-x}{ax\,(x-a)}=\underset{x\to a}{\mathop{\lim }}\,\,\frac{-1}{ax}=\frac{-1}{{{a}^{2}}}\].
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