Question

Solve Limit: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}}$ is equal to
A. $2$
B. $\dfrac{1}{2}$
C. $4$
D. $3$

Answer 1

Here we will solve the given limit by using different identities of trigonometric functions. First, we will try to reduce the terms in the numerator by using the identity of cosine and sine. Then we will use the property of tangent in the limit to get its value. Finally, we will put the value of $x$ and get our desired answer.

Complete step by step solution:
We have to solve:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}}$…..$\left( 1 \right)$
We know the identity $1 - \cos 2x = 2{\sin ^2}x$.
Substituting the above identity in equation $\left( 1 \right)$, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2{{\sin }^2}x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}}$…..$\left( 2 \right)$
Now as we know the identity that the limit of tangent angle divided by the angle as the angle approaches to zero is one i.e. $\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1$.
Now we will multiply and divide the angle of tangent in the equation $\left( 2 \right)$ as,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2{{\sin }^2}x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} \times \dfrac{{4x}}{{4x}}$
Taking the tangent and angle together, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2{{\sin }^2}x} \right)\left( {3 + \cos x} \right)}}{{x \times 4x}} \times \left( {\dfrac{{4x}}{{\tan 4x}}} \right)$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2{{\sin }^2}x} \right)\left( {3 + \cos x} \right)}}{{x \times 4x}} \times 1$
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2{{\sin }^2}x} \right)}}{{x \times 4x}} \times \left( {3 + \cos x} \right)$…..$\left( 3 \right)$
Now, as we know that sine term also follow the same property as tangent for limit of angle tending to 0 which is $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$.
Using above property we can write the equation (3) as,
$\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sin }^2}x}}{{x \times 4x}} \times \left( {3 + \cos x} \right)\\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{2}{4}\left( {\dfrac{{{{\sin }^2}x}}{{{x^2}}}} \right) \times \left( {3 + \cos x} \right)\\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{2}{\left( {\dfrac{{\sin x}}{x}} \right)^2} \times \left( {3 + \cos x} \right)\end{array}$
Using sine property, we get
$\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{1}{2} \times 1 \times \left( {3 + \cos x} \right)\\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{3 + \cos x}}{2}\end{array}$
Now, as the above value can’t be simplified further we will substitute the value of $x$ as 0 and get,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \dfrac{{3 + \cos \left( 0 \right)}}{2}$
As $\cos \left( 0 \right) = 1$, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \dfrac{{3 + 1}}{2}$
Adding the terms in the numerator, we get
$\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = \dfrac{4}{2}\\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = 2\end{array}$

Hence, option (A) is correct.

Note:
Trigonometry is that branch of mathematics which deals with specific functions of angles and also their application in calculations and simplification. The commonly used six types of trigonometry functions are defined as sine, cosine, tangent, cotangent, secant and cosecant. Identities are those equations which are true for every variable.
Limit defines a value that a function approaches. The six trigonometric functions are periodic in nature and therefore don’t approach a finite limit.