The value of \[\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{\left( {1 - \cos x} \right)\left( {1 - \cos 4x} \right)}}{{{x^4}}}\] is equal to:
\[
A.\,\,4 \\
B.\,\,16 \\
C.\,\,32 \\
D.\,\,None\,of\,\,these \\
\]
Answer
594.6k+ views
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.
$ 1 - \cos x = 2{\sin ^2}\left( {\dfrac{x}{2}} \right) $ , $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \theta }}{\theta } = 1 $
Complete step-by-step answer:
Given function is \[\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{\left( {1 - \cos x} \right)\left( {1 - \cos 4x} \right)}}{{{x^4}}}\]
To find its limit we first simplify its numerator by using trigonometric formulas:
We know that
$
\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,
\[
\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{\left[ {2{{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right] \left( {2{{\sin }^2}2x} \right)}}{{{x^4}}} \\
\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{4.{{\sin }^2}\left( {\dfrac{x}{2}} \right){{\sin }^2}2x}}{{{x^4}}} \\
\;
\]
Multiplying and dividing each sine function with its angle.
\[\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{4\dfrac{{{{\sin }^2}\left( {\dfrac{x}{2}} \right)}}{{{{\left( {\dfrac{x}{2}} \right)}^2}}} \times {{\left( {\dfrac{x}{2}} \right)}^2} \times \dfrac{{{{\sin }^2}\left( {2x} \right)}}{{{{\left( {2x} \right)}^2}}} \times {{\left( {2x} \right)}^2}}}{{{x^4}}}\]
$
\Rightarrow \dfrac{{4{x^4}}}{{{x^4}}} \\
\Rightarrow 4 \;
$
Therefore, the required value of limit function is $ 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.
$ 1 - \cos x = 2{\sin ^2}\left( {\dfrac{x}{2}} \right) $ , $ \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \theta }}{\theta } = 1 $
Complete step-by-step answer:
Given function is \[\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{\left( {1 - \cos x} \right)\left( {1 - \cos 4x} \right)}}{{{x^4}}}\]
To find its limit we first simplify its numerator by using trigonometric formulas:
We know that
$
\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,
\[
\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{\left[ {2{{\sin }^2}\left( {\dfrac{x}{2}} \right)} \right] \left( {2{{\sin }^2}2x} \right)}}{{{x^4}}} \\
\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{4.{{\sin }^2}\left( {\dfrac{x}{2}} \right){{\sin }^2}2x}}{{{x^4}}} \\
\;
\]
Multiplying and dividing each sine function with its angle.
\[\mathop {\lim }\limits_{x\xrightarrow{{}}0} \dfrac{{4\dfrac{{{{\sin }^2}\left( {\dfrac{x}{2}} \right)}}{{{{\left( {\dfrac{x}{2}} \right)}^2}}} \times {{\left( {\dfrac{x}{2}} \right)}^2} \times \dfrac{{{{\sin }^2}\left( {2x} \right)}}{{{{\left( {2x} \right)}^2}}} \times {{\left( {2x} \right)}^2}}}{{{x^4}}}\]
$
\Rightarrow \dfrac{{4{x^4}}}{{{x^4}}} \\
\Rightarrow 4 \;
$
Therefore, the required value of limit function is $ 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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

How many of the following diseases can be studied with class 11 biology CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

Which of the following enzymes is used for carboxylation class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

