A) \[0\]
B) \[-1/9\]
C) \[-1/3\]
D) \[3\sqrt{3}\]
Correct Answer: D
Solution :
Since,\[f(x)=\frac{1}{\sqrt{18-{{x}^{2}}}}\] \[\therefore \] \[\underset{x\to 3}{\mathop{\lim }}\,\frac{f(x)-f(3)}{x-3}\] \[=\underset{x\to 3}{\mathop{\lim }}\,\frac{\frac{1}{\sqrt{18-{{x}^{2}}}}-\frac{1}{\sqrt{18-9}}}{x-3}\] \[=\underset{x\to 3}{\mathop{\lim }}\,\frac{\frac{1}{\sqrt{18-{{x}^{2}}}}-\frac{1}{3}}{x-3}\] Applying L' Hospital's rule \[=\underset{x\to 3}{\mathop{\lim }}\,\frac{-\frac{1}{2}{{(18-{{x}^{2}})}^{-3/2}}(-2x)}{1}\] \[={{9}^{-3/2}}(3)=\frac{1}{9}\]
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