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Question: \(f ( 2 ) = 4\) \(f^{'}(2) = 1,\) then \(\lim_{x \rightarrow 2}\frac{xf(2) - 2f(x)}{x - 2} =\)...

f(2)=4f ( 2 ) = 4 f′(2)=1,f^{'}(2) = 1, then lim⁡x→2xf(2)−2f(x)x−2=\lim_{x \rightarrow 2}\frac{xf(2) - 2f(x)}{x - 2} =

A

1

B

2

C

3

D

-2

Answer

2

Explanation

Solution

Given f(2)=4,f′(2)=1f(2) = 4,f^{'}(2) = 1

∴\therefore lim⁡x→2xf(2)−2f(x)x−2=lim⁡x→2xf(2)−2f(2)+2f(2)−2f(x)x−2=\lim_{x \rightarrow 2}\frac{xf(2) - 2f(x)}{x - 2} = \lim_{x \rightarrow 2}\frac{xf(2) - 2f(2) + 2f(2) - 2f(x)}{x - 2} =

lim⁡x→2(x−2)f(2)x−2−lim⁡x→22f(x)−2f(2)x−2\lim_{x \rightarrow 2}\frac{(x - 2)f(2)}{x - 2} - \lim_{x \rightarrow 2}\frac{2f(x) - 2f(2)}{x - 2} = f(2)−2lim⁡x→2f(x)−f(2)x−2f(2) - 2\lim_{x \rightarrow 2}\frac{f(x) - f(2)}{x - 2}

= f(2)−2f′(2)=4−2(1)=4−2=2f(2) - 2f^{'}(2) = 4 - 2(1) = 4 - 2 = 2

Trick : Applying L-Hospital rule, we get lim⁡x→2f(2)−2f′(2)1=2\lim_{x \rightarrow 2}\frac{f(2) - 2f^{'}(2)}{1} = 2.