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Question: The value of \[\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{\left( {1 - \cos x} \right)\left(...

The value of limx0(1cosx)(1cos4x)x4\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{\left( {1 - \cos x} \right)\left( {1 - \cos 4x} \right)}}{{{x^4}}} is equal to:

A.4 B.16 C.32 D.Noneofthese  A.\,\,4 \\\ B.\,\,16 \\\ C.\,\,32 \\\ D.\,\,None\,of\,\,these \\\
Explanation

Solution

Hint : For this type of limit function we first simplify numerator terms by using trigonometric formulas to convert them in term of multiply or divide and then multiplying and dividing each sine or tangent term with its angle to make it one and hence simplifying it to get required value of limit function.
1cosx=2sin2(x2)1 - \cos x = 2{\sin ^2}\left( {\dfrac{x}{2}} \right) , limx0sinθθ=1\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \theta }}{\theta } = 1

Complete step-by-step answer :
Given function is limx0(1cosx)(1cos4x)x4\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{\left( {1 - \cos x} \right)\left( {1 - \cos 4x} \right)}}{{{x^4}}}
To find its limit we first simplify its numerator by using trigonometric formulas:
We know that
cos2x=12sin2x or 2sin2x=1cos2x or 1cosx=2sin2(x2) Also, 1cos4x=2sin22x   \cos 2x = 1 - 2{\sin ^2}x \\\ or \\\ 2{\sin ^2}x = 1 - \cos 2x \\\ or \\\ 1 - \cos x = 2{\sin ^2}\left( {\dfrac{x}{2}} \right) \\\ Also, \\\ 1 - \cos 4x = 2{\sin ^2}2x \;
Using above trigonometric values in the given limit function. We have,

limx0[2sin2(x2)](2sin22x)x4 limx04.sin2(x2)sin22xx4    \mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{\left[ {2{{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right] \left( {2{{\sin }^2}2x} \right)}}{{{x^4}}} \\\ \mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{4.{{\sin }^2}\left( {\dfrac{x}{2}} \right){{\sin }^2}2x}}{{{x^4}}} \\\ \;

Multiplying and dividing each sine function with its angle.
limx04sin2(x2)(x2)2×(x2)2×sin2(2x)(2x)2×(2x)2x4\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{4\dfrac{{{{\sin }^2}\left( {\dfrac{x}{2}} \right)}}{{{{\left( {\dfrac{x}{2}} \right)}^2}}} \times {{\left( {\dfrac{x}{2}} \right)}^2} \times \dfrac{{{{\sin }^2}\left( {2x} \right)}}{{{{\left( {2x} \right)}^2}}} \times {{\left( {2x} \right)}^2}}}{{{x^4}}}
4x4x4 4   \Rightarrow \dfrac{{4{x^4}}}{{{x^4}}} \\\ \Rightarrow 4 \;
Therefore, the required value of limit function is 4.4.
So, the correct answer is “Option A”.

Note : To find the limit of any trigonometric functions we first see if the limit of the function is zero or not. If limit is not zero then we first make it zero by doing some substitution and after or if limit is zero already then we convert trigonometric function into multiplication if in either addition or subtraction by using some trigonometric identities and then simplifying it to get the value of required limit function.