Question

# $$f(x)=\dfrac {1-\cos(1-\cos x)}{x^4}$$ is continuous at $$x= 0$$, then $$f(0)=$$

A
12
B
14
C
16
D
18

Solution

## The correct option is C $$\dfrac{1}{8}$$$$f(x)=\dfrac {1-\cos(1-\cos x)}{x^4}$$It is given that the function $$f$$ is continuous at $$x = 0.$$Then, $$\displaystyle\lim_{x\rightarrow 0}f(x)=f(0)$$$$f(x)=\displaystyle\lim_{x\rightarrow 0}\dfrac {1-\cos(1-\cos x)}{x^4}$$$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{1-\cos\left(2\left(\sin\dfrac{x}{2}\right)^2\right)}{x^4}$$$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{2\left(\sin\left(\sin\dfrac{x}{2}\right)^2\right)^2}{x^4}$$$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{2\left(\sin\left(\sin\dfrac{x}{2}\right)^2\right)^2}{\left(\sin\dfrac{x}{2}\right)^4}\dfrac{\left(\sin\dfrac{x}{2}\right)^4}{x^4}$$$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{2\left(\sin\left(\sin\dfrac{x}{2}\right)^2\right)^2}{\left(\sin\dfrac{x}{2}\right)^4}\dfrac{\left(\sin\dfrac{x}{2}\right)^4}{\left(\dfrac{x}{2}\right)^42^4}$$$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{1}{2^3}\dfrac{\left(\sin\left(\sin\dfrac{x}{2}\right)^2\right)^2}{\left(\sin\dfrac{x}{2}\right)^4}\displaystyle\lim_{x\rightarrow 0}\dfrac{\left(\sin\dfrac{x}{2}\right)^4}{\left(\dfrac{x}{2}\right)^4}$$$$=\dfrac{1}{8}(1)(1)$$$$=\dfrac{1}{8}$$$$\therefore$$  $$f(0)=\dfrac{1}{8}$$Mathematics

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