If $$\left( x \right)$$ is differentiable and strictly increasing function, then the value of $$\mathop {\lim }\limits_{x \to 0} \frac{{f\left( {{x^2}} \right) - f\left( x \right)}}{{f\left( x \right) - f\left( 0 \right)}}$$ is-
A.
$$1$$
B.
$$0$$
C.
$$-1$$
D.
$$2$$
Answer :
$$-1$$
Solution :
$$\eqalign{
& {\text{Let }}L = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( {{x^2}} \right) - f\left( x \right)}}{{f\left( x \right) - f\left( 0 \right)}} \cr
& \left[ {\because f'\left( a \right) > 0,\,f\,{\text{being strictly increasing}}} \right] \cr} $$
Using L'Hospital rule, we get
$$\eqalign{
& L = \mathop {\lim }\limits_{x \to 0} \frac{{f'\left( {{x^2}} \right).2x - f'\left( x \right)}}{{f'\left( x \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{f'\left( {{x^2}} \right).2x}}{{f'\left( x \right)}} - 1 \cr
& = 0 - 1 \cr
& = - 1 \cr} $$
Releted MCQ Question on Calculus >> Differentiability and Differentiation
Releted Question 1
There exist a function $$f\left( x \right),$$ satisfying $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$ for all $$x,$$ and-
A.
$$f''\left( x \right) > 0$$ for all $$x$$
B.
$$ - 1 < f''\left( x \right) < 0$$ for all $$x$$
C.
$$ - 2 \leqslant f''\left( x \right) \leqslant - 1$$ for all $$x$$
If $$f\left( a \right) = 2,\,f'\left( a \right) = 1,\,g\left( a \right) = - 1,\,g'\left( a \right) = 2,$$ then the value of $$\mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}$$ is-
Let $$f:R \to R$$ be a differentiable function and $$f\left( 1 \right) = 4.$$ Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$ is-