Question

# The value of $$f(0)$$ so that the function $$\mathrm{f}({x})$$$$=\displaystyle \frac{1-\cos(1-\cos x)}{x^{4}}$$ is continuous everywhere is

A
18
B
12
C
14
D
13

Solution

## The correct option is A $$\displaystyle \frac{1}{8}$$$$f\left( x \right) =\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) }$$$$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x } \right) } }{ { x }^{ 4 } } }$$$$=\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x } \right) } }{ { (1-\cos { x } ) }^{ 2 } } } \times \cfrac { { (1-\cos { x } ) }^{ 2 } }{ { x }^{ 4 } }$$$$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x } \right) } }{ { (1-\cos { x } ) }^{ 2 } } } \times \displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cfrac { 1-\cos { x } }{ { x }^{ 2 } } \right) ^{ 2 } } }$$As $$x\rightarrow 0$$$$1-\cos { x } \rightarrow 0$$$$=\displaystyle \lim _{ 1-\cos { x } \rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x } \right) } }{ { (1-\cos { x } ) }^{ 2 } } } \times { \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }$$$$=\cfrac { 1 }{ 2 } \times { \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }=\cfrac { 1 }{ 8 }$$Mathematics

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