Question

Evaluate $\mathop {\lim }\limits_{x \to 0} \cos x\cot 2x$.

Answer 1

Hint: In this question, we are given a trigonometric equation and we have to find the value of its limit when the limit tends to $0$. Use certain basic trigonometric formulae to first expand it and then use L’Hospital’s rule to solve the indeterminate form. Lastly, put $x = 0$.

Formula used: 1) $\cot x = \dfrac{{\cos x}}{{\sin x}}$
2) $\sin 2x = 2\sin x\cos x$

Complete step-by-step solution:
We have a trigonometric equation and we have to find the value of its limit when the limit tends to $0$ . But we cannot directly put $x = 0$. So, we will first simplify the equations by using certain very basic trigonometric formulae.
To find: $\mathop {\lim }\limits_{x \to 0} \cos x\cot 2x$
First, we will put $\cot x = \dfrac{{\cos x}}{{\sin x}}$ .
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \cos x\dfrac{{\cos 2x}}{{\sin 2x}}$
Now, we will simplify the denominator by putting $\sin 2x = 2\sin x\cos x$.
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x\cos 2x}}{{2\sin x\cos x}}$
In the next step, we will cancel out $\cos x$ as it is present in both numerator and denominator. We will get,
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos 2x}}{{2\sin x}}$
We will take out the constant from the denominator now.
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{1}{2} \times \dfrac{{\cos 2x}}{{\sin x}}$
$ \Rightarrow \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos 2x}}{{\sin x}}$
In this, we will use L’Hospital’s rule. This rule says –
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
Using the rule,
$ \Rightarrow \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{\cos 2x}}{{\sin x}} = \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{ - 2\sin 2x}}{{\cos x}}$
Now, we will put limits $x = 0$.
$ \Rightarrow \dfrac{1}{2}\mathop {\lim }\limits_{x \to 0} \dfrac{{ - 2\sin 2x}}{{\cos x}} = \dfrac{1}{2} \times \dfrac{{ - 2\sin 0^\circ }}{{\cos 0^\circ }}$
Putting the values,
$ \Rightarrow \dfrac{1}{2} \times \dfrac{0}{1}$

Hence, the answer is $0$.

Note: L’Hospital’s Rule:
This rule is basically a technique which is used to evaluate limits of indeterminate forms. The rule often converts an indeterminate form to an expression that can be easily evaluated by substitution. It basically stated that the limit of an indeterminate function is equal to the limit of their differentiation. The differentiation of the numerator and denominator often simplifies the quotient or converts it to a limit that can be evaluated directly. There is no limit on differentiating the question. You can differentiate it as many times as possible or till you get the most simplified value to get the required answer.
$ \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$