A) 2
B) \[-2\]
C) \[-4\]
D) 3
Correct Answer: C
Solution :
\[\underset{x\to 2}{\mathop{\lim }}\,\frac{xf(2)-2f(x)}{x-2}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{xf(2)-2f(2)+2f(2)-2f(x)}{x-2}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{f(2)(x-2)-2\{f(x)-f(2)\}}{x-2}\] \[=f(2)-2\underset{x\to 2}{\mathop{\lim }}\,\,\frac{f(x)-f(2)}{x-2}\] \[=f(2)-2f\,'(2)=4-2\times 4=-4\] Alternate Solution \[\underset{x\to 2}{\mathop{\lim }}\,\left\{ \frac{xf(2)-2f(x)}{x-2} \right\}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\{f(2)-2f'(x)\}\] (by L? Hospital rule) \[=f(2)-2f'(2)=4-2\times 4=-4\]
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