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If$f(a)=2;f '(a)=1;(g)=(a)=-1;g '(a)=2$ then the value of $\;\lim \limits_{x \to 2}\Large\frac{g(x)f(a)-g(a)f(x)}{x-a}$ is

\[\begin{array}{1 1}(1)5&(2)-5\\(3)3&(4)-3\end{array}\]

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$\lim \limits _{x \to a} \large\frac{g(x)f(a)-g(a)f(x)}{x-a}=\frac{g(a)f(a)-g(a)f(a)}{a-a}$
$\qquad= \large\frac{0}{0}$
Which is a indeterminate
By L'Hospital Rule
$\lim \limits _{x \to a} \large\frac{g(x)f(a)-g(a)f(x)}{x-a}$
$\qquad=\lim \limits _{x \to a} \large\frac{g'(x)f(a)-g(a)f'(x)}{1}$
$\qquad= g'(a)f(a) -g(a) f'(a)$
$\qquad= 2 \times 2 -(-1) (1)$
$\qquad=4+1=5$
Hence 1 is the correct answer.
answered May 20, 2014 by meena.p
 

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