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# 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) $-\dfrac{1}{4}$

Last updated date: 13th Sep 2024
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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.

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: $\displaystyle \lim_{x \to 0}\dfrac{\text{cose}{{\text{c}}^{2}}x}{\left( {{\cot }^{2}}x \right)\left( 4-\sec x \right)}$ .