CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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



A
12
loader
B
14
loader
C
16
loader
D
18
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image