If f(2)=4 and f1(2)=1 then Ltx→2x f(2)-2 f(x)x-2=

f(2)=4, f1(2)=1 ltx→2 xf(2)-2f(x)x-2 00 form =ltx→2 f(2)-2f1(x)1 (L' Hospital Rule)=f(2)-2f1(2) =4-2(1) =2

If f(2)=4 and f1(2)=1 then Ltx→2x f(2)-2 f(x)x-2=

f(2)=4, f1(2)=1 ltx→2 xf(2)-2f(x)x-2 00 form =ltx→2 f(2)-2f1(x)1 (L' Hospital Rule)=f(2)-2f1(2) =4-2(1) =2

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If $f (2) = 4$ and $f^{1} (2) = 1 then$ $\underset{x \to 2}{L t} \frac{x f (2) - 2 f (x)}{x - 2} =$

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Detailed Solution

$f (2) = 4, f^{1} (2) = 1$
$\underset{x \to 2}{lt} \frac{xf (2) - 2 f (x)}{x - 2} (\frac{0}{0} form) = \underset{x \to 2}{lt} \frac{f (2) - 2 f^{1} (x)}{1} (L' Hospital Rule)$
$= f (2) - 2 f^{1} (2) = 4 - 2 (1) = 2$

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If $f (2) = 4$ and $f^{1} (2) = 1 then$ $\underset{x \to 2}{L t} \frac{x f (2) - 2 f (x)}{x - 2} =$