Let $$f''\left( x \right)$$  be continuous at $$x = 0$$  and $$f''\left( 0 \right) = 4.$$
Then value of $$\mathop {\lim }\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}}$$       is :

Solution :
$$\eqalign{ & {\text{Given }}f''\left( x \right){\text{ is continuous at }}x = 0 \cr & = \mathop {\lim }\limits_{x \to 0} f''\left( x \right) = f''\left( 0 \right) = 4 \cr & {\text{Now, }}\mathop {\lim }\limits_{x \to 0} \frac{{2f\left( x \right) - 3f\left( {2x} \right) + f\left( {4x} \right)}}{{{x^2}}}\,\,\,\,\left[ {\frac{0}{0}{\text{ form}}} \right] \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2f'\left( x \right) - 6f'\left( {2x} \right) + 4f'\left( {4x} \right)}}{{2x}}\,\,\,\,\left[ {\frac{0}{0}{\text{ form}}} \right] \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2f''\left( x \right) - 12f''\left( x \right) + 16f''\left( {4x} \right)}}{2} \cr & \left[ {{\text{Using L'Hospital rule successively}}} \right] \cr & = \frac{{2f''\left( 0 \right) - 12f''\left( 0 \right) + 16f''\left( 0 \right)}}{2} \cr & = 12 \cr} $$