Question

Evaluate the following limit:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cot 2x - \cos ec2x} \right)}}{x}$

Answer 1

Hint : To find the value of the limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cot 2x - \cos ec2x} \right)}}{x}$ , first of all substitute $\cot 2x = \dfrac{{\cos 2x}}{{\sin 2x}}$ and $\cos ec2x = \dfrac{1}{{\sin 2x}}$ . Then simplify and take – sign common and use the formula $1 - \cos 2x = 2{\sin ^2}x$ . Now, we will get $\dfrac{{\sin x}}{x} = 1$ . Then, use the formula $\sin 2x = 2\sin x\cos x$ and now we can put the value of the limit that is $x = 0$ and find the answer.

Complete step-by-step answer :
In this question, we are asked to find the value of the given limit.
Given limit: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cot 2x - \cos ec2x} \right)}}{x}$ - - - - - - - - - - - (1)
First of all, let this limit be equal to L. Therefore,
$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cot 2x - \cos ec2x} \right)}}{x}$
Now, we know that cot is the ratio of cos and sin and cosec is the inverse of sin. So, we can say that
$\Rightarrow \cot 2x = \dfrac{{\cos 2x}}{{\sin 2x}}$
And
$\Rightarrow \cos ec2x = \dfrac{1}{{\sin 2x}}$
Therefore, substituting these values in equation (1), we get
$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{{\cos 2x}}{{\sin 2x}} - \dfrac{1}{{\sin 2x}}} \right)}}{x}$
Now, taking LCM, we get
$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( {\left( {\dfrac{{\cos 2x - 1}}{{\sin 2x}}} \right)\dfrac{1}{x}} \right)$
Now, taking out minus sign (-), we get
$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \left( {\dfrac{{1 - \cos 2x}}{{\sin 2x}}} \right)\dfrac{1}{x}} \right)$ - - - - - - - - - - - - - (2)
Now, we know the trigonometric relation that,
$\Rightarrow 1 - \cos 2x = 2{\sin ^2}x$
Therefore, substituting this value in equation (2), we get
$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \left( {\dfrac{{2{{\sin }^2}x}}{{\sin 2x}}} \right)\dfrac{1}{x}} \right)$ - - - - - - - - (3)
$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \dfrac{{\sin x}}{x}\left( {\dfrac{{2\sin x}}{{\sin 2x}}} \right)} \right)$
Now, according to the rule of limits,
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \theta }}{\theta } = 1$
Therefore, we get
$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \left( 1 \right)\left( {\dfrac{{2\sin x}}{{\sin 2x}}} \right)} \right)$
Now, we have the formula
$\Rightarrow \sin 2x = 2\sin x\cos x$
Therefore, we get

$$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \left( 1 \right)\left( {\dfrac{{2\sin x}}{{2\sin x\cos x}}} \right)} \right) \\\ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \left( {\dfrac{1}{{\cos x}}} \right)} \right) \\\ \Rightarrow L = - \left( {\dfrac{1}{{\cos 0}}} \right) \\\ \Rightarrow L = - \left( {\dfrac{1}{1}} \right) \\\ \Rightarrow L = - 1 \;$$

Hence, $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cot 2x - \cos ec2x} \right)}}{x}$ is equal to -1.
So, the correct answer is “-1”.

Note : Here, we can also find our answer without putting the value of the limit. Here, by equation (3), we have
$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \left( {\dfrac{{2{{\sin }^2}x}}{{\sin 2x}}} \right)\dfrac{1}{x}} \right)$
Now, $\sin 2x = 2\sin x\cos x$ . Therefore, we get
$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \left( {\dfrac{{2{{\sin }^2}x}}{{2\sin x\cos x}}} \right)\dfrac{1}{x}} \right) \\\ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)\dfrac{1}{x}} \right) \\\ \Rightarrow L = \mathop {\lim }\limits_{x \to 0} \left( { - \dfrac{{\tan x}}{x}} \right) \;$
Now, we know that
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan \theta }}{\theta } = 1$
Therefore,
$\Rightarrow L = - 1$