Question

Find $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos \left( {1 - \cos x} \right)}}{{{x^4}}}$?

Answer 1

Hint:
This type of problem can be solved in various ways. It can be directly solved by manipulating the equations by dividing and multiplying with ${\left( {1 - \cos x} \right)^2}$ and then solving for it by using the condition given to us. Another method is by using the formula $\cos 2x = 1 - 2{\sin ^2}x$ and from here we will get the value of $1 - \cos x$and can easily solve them.

Formula used:
Trigonometric formulas used in this question will be:
$\cos 2x = 1 - 2{\sin ^2}x$

Complete step by step solution:
 As we know the formula
$ \Rightarrow \cos 2x = 1 - 2{\sin ^2}x$
So it can also be written as
$ \Rightarrow 2{\sin ^2}x = 1 - \cos 2x$
And also it can be written as
$ \Rightarrow 1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$
So we have the question which is,
$\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos \left( {1 - \cos x} \right)}}{{{x^4}}}$
Therefore, by using the equation and substituting the values, we get
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} 2{\sin ^2}\dfrac{{\left( {1 - \cos x} \right)}}{2}\left( {\dfrac{1}{{{x^4}}}} \right)$
On further solving more, we get
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} 2{\sin ^2}\dfrac{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}}{{{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}^2}}}\left( {\dfrac{1}{{{x^4}}}} \right){\left( {{{\sin }^2}\dfrac{x}{2}} \right)^2}$
Now taking the constant term outside, and solving the limit we get
$ \Rightarrow 2\mathop {\lim }\limits_{x \to 0} {\sin ^2}\dfrac{{\sin \left( {{{\sin }^2}\dfrac{x}{2}} \right)}}{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}}\dfrac{{\left( {{{\sin }^4}\dfrac{x}{2}} \right)}}{{\dfrac{{{x^4}}}{{{2^4}}} \cdot {2^4}}}$
Now we will put the limit and solve it, we get
$ \Rightarrow 2{\left( 1 \right)^2}\mathop {\lim }\limits_{x \to 0} {\left( {\dfrac{{\left( {\sin \dfrac{x}{2}} \right)}}{{\dfrac{x}{2}}}} \right)^4}\dfrac{1}{{16}}$
And since the constant term will be reduced and also we will solve for the limit, we get
$ \Rightarrow \dfrac{1}{8}{\left( 1 \right)^4} = \dfrac{1}{8}$

Therefore, $\dfrac{1}{8}$will be the correct answer.

Note:
There is one more method through which this question can be solved and in this method, we don’t have to remember any properties or trigonometric identities to solve it. It can be solved by multiplying and dividing the question ${\left( {1 - \cos x} \right)^2}$. And the equation will look like $\mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos \left( {1 - \cos x} \right)}}{{{{\left( {1 - \cos x} \right)}^2}}} \times \dfrac{{{{\left( {1 - \cos x} \right)}^2}}}{{{x^4}}}$ and then solving for it with the limit given as $1 - \cos x \to 0$. This will also become zero. And then we will get the result by solving the easy maths. There is one more method by using the expansion of cosine terms we can have the result. So these are the ways through which we can solve this question.