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  • Find the value of <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos (1-\cos x)))}{x^8}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightarrow

Find the value of \mathrm{\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos (1-\cos x)))}{x^8}}
 

Option: 1

\frac{1}{2}


Option: 2

0


Option: 3

1


Option: 4

-1


Answers (1)

best_answer

In this problem, we can take help of \mathrm{\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2}}, which is very easy to establish.

\mathrm{\lim _{x \rightarrow 0} \frac{(1-\cos x)}{x^2} \times \frac{(1+\cos x)}{(1+\cos x)} =\lim _{x \rightarrow 0} \frac{1-\cos ^2 x}{x^2(1+\cos x)} }

\mathrm{ =\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2(1+\cos x)} }

\mathrm{ =\frac{1}{2}\left(\text { As } \lim _{x \rightarrow 10} \frac{\sin ^2 x}{x^2}=1 \text { and } \lim _{x \rightarrow 0} \frac{1}{1+\cos x}=\frac{1}{2}\right)}

Now, this limit can be generalized as

\mathrm{ \lim _{f(x)=0} \frac{1-\cos (f(x))}{f^2(x)}=\frac{1}{2} }

Hence option 1 is correct.
 

Posted by

jitender.kumar

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