Question

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

Answer 1

Hint: Use the formula $\cos 2x=1-2{{\sin }^{2}}x$ and simplify the expression you get. Then divide the denominator and numerator by x and use the formulae $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1\text{ and }\displaystyle \lim_{x \to 0}\dfrac{\tan kx}{x}=k$ , where k is a constant, to further simplify and reach to the final answer.

Complete step-by-step answer:
In the expression given in the question, if we put the limit and check, then we will find that the expression turns out to be of the form $\dfrac{0}{0}$, which is an indeterminate form.
So, using the formula $\cos 2x=1-2{{\sin }^{2}}x$ in our expression, we get
$\displaystyle \lim_{x \to 0}\dfrac{\left( 1-\cos 2x \right)\left( 3+\cos x \right)}{x\tan 4x}$
$=\displaystyle \lim_{x \to 0}\dfrac{\left( 1-\left( 1-2{{\sin }^{2}}x \right) \right)\left( 3+\cos x \right)}{x\tan 4x}$
$=\displaystyle \lim_{x \to 0}\dfrac{2{{\sin }^{2}}x\left( 3+\cos x \right)}{x\tan 4x}$
Now, we will divide the numerator and denominator by x. On doing so, we get
$=\displaystyle \lim_{x \to 0}\dfrac{2{{\sin }^{2}}x\left( 3+\cos x \right)}{x\times \dfrac{x\tan 4x}{x}}$
Now, we will rearrange the terms to convert the terms to suitable forms. On doing so, we get
$=\displaystyle \lim_{x \to 0}2\times \dfrac{{{\sin }^{2}}x}{{{x}^{2}}}\times \dfrac{1}{\dfrac{\tan 4x}{x}}\times \left( 3+\cos x \right)$
$=\displaystyle \lim_{x \to 0}2\times {{\left( \dfrac{\sin x}{x} \right)}^{2}}\times \dfrac{1}{\dfrac{\tan 4x}{x}}\times \left( 3+\cos x \right)$
Now, we will use the formulae $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1\text{ and }\displaystyle \lim_{x \to 0}\dfrac{\tan kx}{x}=k$ , where k is a constant, to further simplify. On doing so, we get
 $=\displaystyle \lim_{x \to 0}2\times {{1}^{2}}\times \dfrac{1}{4}\times \left( 3+\cos x \right)$
$=\displaystyle \lim_{x \to 0}\dfrac{\left( 3+\cos x \right)}{2}$
Now, we will put the limit and we know that $\cos 0{}^\circ =1$ .
$=\dfrac{3+\cos 0{}^\circ }{2}=\dfrac{3+1}{2}=2$
Therefore, the answer to the above question is 2.

So, the correct answer is “Option c”.

Note: Whenever you come across the forms $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ and cannot think of a formulae to use always give a try to l-hospital’s rule as this may give you the answer. Also, keep a habit of checking the indeterminate forms of the expressions before starting with the questions of limit as this helps you to select the shortest possible way to reach the answer.