CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
12
loader
C
14
loader
D
13
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image