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

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 $