If \[f(2)=4\], \[f'(2)=1\]then \[\underset{x\to 2}{\mathop{\lim }}\,\frac{xf(2)-2f(x)}{x-2}=\] [RPET 1995, 2000]

If \[f(2)=4\], \[f'(2)=1\]then \[\underset{x\to 2}{\mathop{\lim }}\,\f

1.	If \[f(2)=4\], \[f'(2)=1\]then \[\underset{x\to 2}{\mathop{\lim }}\,\frac{xf(2)-2f(x)}{x-2}=\] [RPET 1995, 2000]
A.	1
B.	2
C.	3
D.	?2
Answer» C. 3

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If \[f(2)=4\], \[f'(2)=1\]then \[\underset{x\to 2}{\mathop{\lim }}\,\frac{xf(2)-2f(x)}{x-2}=\] [RPET 1995, 2000]

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