Question

Find the value of the given limit: $ \displaystyle \lim_{x \to 0}\dfrac{\left( 1-\cos 2x \right)\left( 3+\cos x \right)}{x\tan 4x} $ ?
(a) $ \dfrac{1}{2} $
(b) 1
(c) 2
(d) $ -\dfrac{1}{4} $

Answer 1

Hint: We start solving the problem by equating the given limit to L. We then make use of the result $ 1-\cos ax=2{{\sin }^{2}}\left( \dfrac{ax}{2} \right) $ to proceed through the problem. We then make the necessary arrangements in the problem and make use of the result $ \displaystyle \lim_{x \to a}\left( pq \right)=\displaystyle \lim_{x \to a}\left( p \right)\times \displaystyle \lim_{x \to a}\left( q \right) $ to further through the problem. We then make the necessary calculations and use the results $ \displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1 $ and $ \displaystyle \lim_{x \to 0}\dfrac{\tan ax}{ax}=1 $ to get the required value of the given limit in the problem.

Complete step by step answer:
According to the problem, we are asked to find the value of the given limit: $ \displaystyle \lim_{x \to 0}\dfrac{\left( 1-\cos 2x \right)\left( 3+\cos x \right)}{x\tan 4x} $ .
Let us assume $ L=\displaystyle \lim_{x \to 0}\dfrac{\left( 1-\cos 2x \right)\left( 3+\cos x \right)}{x\tan 4x} $ ---(1).
We know that $ 1-\cos ax=2{{\sin }^{2}}\left( \dfrac{ax}{2} \right) $ . Let us use this result in equation (1).
 $ \Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{\left( 2{{\sin }^{2}}x \right)\left( 3+\cos x \right)}{x\tan 4x} $ ---(2).
Now, let us multiply the numerator and denominator of the limit in equation (2) with x.
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{\left( 2{{\sin }^{2}}x \right)\left( 3+\cos x \right)}{x\tan 4x}\times \dfrac{x}{x}\].
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\dfrac{\left( 2x{{\sin }^{2}}x \right)\left( 3+\cos x \right)}{{{x}^{2}}\tan 4x}\] ---(3).
We know that $ \displaystyle \lim_{x \to a}\left( pq \right)=\displaystyle \lim_{x \to a}\left( p \right)\times \displaystyle \lim_{x \to a}\left( q \right) $ . Let us use this result in equation (3).
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\left( \dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \right)\times \displaystyle \lim_{x \to 0}\left( 3+\cos x \right)\times \displaystyle \lim_{x \to 0}\dfrac{2x}{\tan 4x}\].
\[\Rightarrow L=\displaystyle \lim_{x \to 0}\left( \dfrac{\sin x}{x} \right)\times \displaystyle \lim_{x \to 0}\left( \dfrac{\sin x}{x} \right)\times \displaystyle \lim_{x \to 0}\left( 3+\cos x \right)\times \displaystyle \lim_{x \to 0}\dfrac{1}{2\times \dfrac{\tan 4x}{4x}}\] ---(4).
We know that $ \displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1 $ and $ \displaystyle \lim_{x \to 0}\dfrac{\tan ax}{ax}=1 $ . Let us use this results in equation (4).
\[\Rightarrow L=\left( 1 \right)\times \left( 1 \right)\times \left( 3+\cos 0 \right)\times \dfrac{1}{2\times 1}\].
\[\Rightarrow L=\left( 3+1 \right)\times \dfrac{1}{2}\].
\[\Rightarrow L=4\times \dfrac{1}{2}\].
\[\Rightarrow L=2\].
So, we have found the value of the given limit as 2.
 $ \therefore, $ The correct option for the given problem is (c).

Note:
We should not make calculation mistakes while solving this problem. We should not confuse $ \displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1 $ with $ \displaystyle \lim_{x \to 0}\dfrac{\cos x}{x}=1 $ which is the common mistake done by students. We can also make use of the L-Hospital rule as we can see that the limits tend to give indeterminate form $ \left( \dfrac{0}{0} \right) $ on applying the limit directly. Similarly, we can expect problems to find the value of the limit: $ \displaystyle \lim_{x \to 0}\dfrac{\text{cose}{{\text{c}}^{2}}x}{\left( {{\cot }^{2}}x \right)\left( 4-\sec x \right)} $ .