If $$f\left( x \right),\,g\left( x \right)$$ be differentiable functions and $$f\left( 1 \right) = g\left( 1 \right) = 2,$$ then $$\mathop {\lim }\limits_{x \to 1} \frac{{f\left( 1 \right)g\left( x \right) - f\left( x \right)g\left( 1 \right) - f\left( 1 \right) + g\left( 1 \right)}}{{g\left( x \right) - f\left( x \right)}}$$ is equal to :
A.
0
B.
1
C.
2
D.
none of these
Answer :
2
Solution :
$$\eqalign{
& {\text{Limit}} = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( 1 \right)g'\left( x \right) - f'\left( x \right)g\left( 1 \right)}}{{g'\left( x \right) - f'\left( x \right)}} \cr
& = \mathop {\lim }\limits_{x \to 1} \frac{{2\left\{ {g'\left( x \right) - f'\left( x \right)} \right\}}}{{g'\left( x \right) - f'\left( x \right)}} \cr
& = \mathop {\lim }\limits_{x \to 1} 2 \cr
& = 2 \cr} $$
Releted MCQ Question on Calculus >> Limits
Releted Question 1
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If $$\eqalign{
& f\left( x \right) = \frac{{\sin \left[ x \right]}}{{\left[ x \right]}},\,\,\left[ x \right] \ne 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ x \right] = 0 \cr} $$
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