Question

Find the value of ${\lim }_{x \to 0} \dfrac{{\left[ {1 - \cos \left( {1 - \cos x} \right)} \right]}}{{{x^4}}}$.
A. $\dfrac{1}{2}$
B. $\dfrac{1}{4}$
C. $\dfrac{1}{6}$
D. $\dfrac{1}{8}$

Answer 1

Hint: The given problem is based on the limits in calculus. So here, we first apply the formula $1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$ in the given equation. After that, we rewrite the expression in the form sine function and change the limit to apply the formula to get the value of the equation

Formula Used:
$1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$
${\lim }_{x \to a} f\left( x \right) \cdot g\left( x \right) = {\lim }_{x \to a} f\left( x \right) \cdot {\lim }_{x \to a} g\left( x \right)$
${\lim }_{x \to a} \dfrac{{\sin \left( {x - a} \right)}}{{x - a}} = 1$
${\lim }_{x \to a} k = k$ where $k$ is a constant.

Complete step by step solution:
Given limit is ${\lim }_{x \to 0} \dfrac{{\left[ {1 - \cos \left( {1 - \cos x} \right)} \right]}}{{{x^4}}}$.
Now we will apply the formula $1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$.
$ = {\lim }_{x \to 0} \dfrac{{\left[ {1 - \cos \left( {2{{\sin }^2}\dfrac{x}{2}} \right)} \right]}}{{{x^4}}}$
Again, apply the formula $1 - \cos 2x = 2{\sin ^2}x$
$ = {\lim }_{x \to 0} \dfrac{{\left[ {2{{\sin }^2}\left( {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{2}} \right)} \right]}}{{{x^4}}}$
$ = {\lim }_{x \to 0} \dfrac{{\left[ {2{{\sin }^2}\left( {{{\sin }^2}\dfrac{x}{2}} \right)} \right]}}{{{x^4}}}$
Now multiply ${\left( {{{\sin }^2}\dfrac{x}{2}} \right)^2}$ in the numerator and denominator

$ = 2{\lim }_{x \to 0} \left[ {\dfrac{{{{\sin }^2}\left( {{{\sin }^2}\dfrac{x}{2}} \right)}}{{{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}^2}}} \cdot \dfrac{{{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}^2}}}{{{x^4}}}} \right]$
Rewrite in the form of $\dfrac{{\sin x}}{x}$

$ = 2{\lim }_{x \to 0} \left[ {\dfrac{{{{\sin }^2}\left( {{{\sin }^2}\dfrac{x}{2}} \right)}}{{{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}^2}}} \cdot \dfrac{{{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}^2}}}{{\dfrac{{{x^4}}}{{{2^4}}} \cdot {2^4}}}} \right]$
$ = 2{\lim }_{x \to 0} \left[ {\dfrac{{{{\sin }^2}\left( {{{\sin }^2}\dfrac{x}{2}} \right)}}{{{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}^2}}} \cdot \dfrac{{{{\left( {\sin \dfrac{x}{2}} \right)}^4}}}{{\dfrac{{{x^4}}}{{{2^4}}}}} \cdot \dfrac{1}{{{2^4}}}} \right]$
Now we apply the formula ${\lim }_{x \to a} f\left( x \right) \cdot g\left( x \right) = {\lim }_{x \to a} f\left( x \right) \cdot {\lim }_{x \to a} g\left( x \right)$
$ = 2{\lim }_{x \to 0} \dfrac{{{{\sin }^2}\left( {{{\sin }^2}\dfrac{x}{2}} \right)}}{{{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}^2}}} \cdot {\lim }_{x \to 0} \dfrac{{{{\left( {\sin \dfrac{x}{2}} \right)}^4}}}{{\dfrac{{{x^4}}}{{{2^4}}}}} \cdot {\lim }_{x \to 0} \dfrac{1}{{{2^4}}}$

Since $x \to 0$
$ \Rightarrow \dfrac{x}{2} \to 0$
$ \Rightarrow \sin \dfrac{x}{2} \to 0$
$ \Rightarrow {\sin ^2}\dfrac{x}{2} \to 0$
Now changing the limits
$ = 2 \cdot {\lim }_{{{\sin }^2}\dfrac{x}{2} \to 0} {\left( {\dfrac{{\sin \left( {{{\sin }^2}\dfrac{x}{2}} \right)}}{{{{\sin }^2}\dfrac{x}{2}}}} \right)^2} \cdot {\lim }_{\dfrac{x}{2} \to 0} {\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}} \right)^4} \cdot {\lim }_{x \to 0} \dfrac{1}{{{2^4}}}$
Now we apply the formula ${\lim }_{x \to a} \dfrac{{\sin \left( {x - a} \right)}}{{x - a}} = 1$ and ${\lim }_{x \to a} k = k$
$ \Rightarrow 2 \cdot {1^2} \cdot {1^4} \cdot \dfrac{1}{{{2^4}}}$
$ \Rightarrow \dfrac{1}{{{2^3}}}$
$ \Rightarrow \dfrac{1}{8}$

Option ‘D’ is correct

Note: For solving this we are going to use a set of differentiation formulas and L-Hospital’s rule. A common mistake that we make here is not writing the expression in the form $\dfrac{{\sin x}}{x}$. And solving it directly in the form $2{\lim }_{x \to 0} \dfrac{{{{\sin }^2}\left( {{{\sin }^2}\dfrac{x}{2}} \right)}}{{{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}^2}}} \cdot {\lim }_{x \to 0} \dfrac{{{{\left( {{{\sin }^2}\dfrac{x}{2}} \right)}^2}}}{{{x^4}}}$ and apply the formula ${\lim }_{x \to a} \dfrac{{\sin \left( {x - a} \right)}}{{x - a}} = 1$ which is incorrect method.