Question

Let $$f\left( x \right)$$  be a twice-differentiable function and $$f''\left( 0 \right) = 2,$$   then $$\mathop {\lim }\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}}$$      is :

A. 6  
B. 3
C. 12
D. none of these
Answer :   6
Solution :
$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{2f'\left( x \right) - 6f'\left( {2x} \right) + 4f'\left( {4x} \right)}}{{2x}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2f''\left( x \right) - 12f''\left( {2x} \right) + 16f''\left( {4x} \right)}}{2} \cr & = \frac{{6f''\left( 0 \right)}}{2} \cr & = 6 \cr} $$

Releted MCQ Question on
Calculus >> Limits

Releted Question 1

lf $$f\left( x \right) = \sqrt {\frac{{x - \sin \,x}}{{x + {{\cos }^2}x}}} ,$$     then $$\mathop {\lim }\limits_{x\, \to \,\infty } f\left( x \right)$$    is-

A. $$0$$
B. $$\infty $$
C. $$1$$
D. none of these
Releted Question 2

If $$G\left( x \right) = - \sqrt {25 - {x^2}} $$     then $$\mathop {\lim }\limits_{x\, \to \,{\text{I}}} \frac{{G\left( x \right) - G\left( I \right)}}{{x - 1}}$$     has the value-

A. $$\frac{1}{{24}}$$
B. $$\frac{1}{{5}}$$
C. $$ - \sqrt {24} $$
D. none of these
Releted Question 3

$$\mathop {\lim }\limits_{n\, \to \,\infty } \left\{ {\frac{1}{{1 - {n^2}}} + \frac{2}{{1 - {n^2}}} + ..... + \frac{n}{{1 - {n^2}}}} \right\}$$        is equal to-

A. $$0$$
B. $$ - \frac{1}{2}$$
C. $$ \frac{1}{2}$$
D. none of these
Releted Question 4

If $$\eqalign{ & f\left( x \right) = \frac{{\sin \left[ x \right]}}{{\left[ x \right]}},\,\,\left[ x \right] \ne 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ x \right] = 0 \cr} $$
Where \[\left[ x \right]\] denotes the greatest integer less than or equal to $$x.$$ then $$\mathop {\lim }\limits_{x\, \to \,0} f\left( x \right)$$   equals

A. $$1$$
B. $$0$$
C. $$ - 1$$
D. none of these

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Limits


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