A) \[\frac{2}{9}\]
B) \[-\frac{2}{49}\]
C) \[\frac{1}{56}\]
D) \[-\frac{1}{56}\]
Correct Answer: D
Solution :
Applying L-Hospital?s rule, \[\underset{x\to 7}{\mathop{\lim }}\,\frac{2-\sqrt{x-3}}{{{x}^{2}}-49}\] = \[\underset{x\to 7}{\mathop{\lim }}\,\frac{0-\frac{1}{2\sqrt{x-3}}}{2x}\] \[=\underset{x\to 7}{\mathop{\lim }}\,\frac{-1}{4x\sqrt{x-3}}=\frac{-1}{4.7\sqrt{7-3}}=\frac{-1}{56}\].
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