Question

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x\cot \left( 4x \right)}{{{\sin }^{2}}x{{\cot }^{2}}\left( 2x \right)}$is equal to:
(a) 2
(b) 0
(c) 4
(d) 1

Answer 1

Hint: First, before proceeding for this, we must go with the conventional method as by substituting the value of x as 0 in the given question. Then, we get the answer in the form which is not a valid answer and to simplify we use the trigonometric identity as $\cot x=\dfrac{1}{\tan x}$. Then, by using the formula for the limit as $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1,\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\tan x}{x}=1$, we get the final result.

Complete step by step answer:
In this question, we are supposed to find the value of the limit as $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x\cot \left( 4x \right)}{{{\sin }^{2}}x{{\cot }^{2}}\left( 2x \right)}$.
So, before proceeding for this, we must go with the conventional method as by substituting the value of x as 0 in the given question as:
$\begin{align}
  & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{x\cot \left( 4x \right)}{{{\sin }^{2}}x{{\cot }^{2}}\left( 2x \right)}=\dfrac{0\times \cot \left( 0 \right)}{{{\sin }^{2}}\left( 0 \right){{\cot }^{2}}\left( 0 \right)} \\
 & \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{x\cot \left( 4x \right)}{{{\sin }^{2}}x{{\cot }^{2}}\left( 2x \right)}=\dfrac{0\times \infty }{0\times \infty } \\
\end{align}$
Then, we get the answer in the form which is not a valid answer and to simplify we use the trigonometric identity as:
$\cot x=\dfrac{1}{\tan x}$
So, by using the above mentioned identity, we get the limit function as:
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{x}{\tan \left( 4x \right)}}{\dfrac{{{\sin }^{2}}x}{{{\tan }^{2}}\left( 2x \right)}}$
Then, by using the formula for the limit as:
$\begin{align}
  & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1 \\
 & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\tan x}{x}=1 \\
\end{align}$
Then, by using the above stated formulas and also we can make the functions as sine and tangent function in question as required to get the limits solved.
So, b y multiplying and dividing the number 4 in the numerator and similarly multiplying the term $4{{x}^{2}}$, we get:
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{4x}{4\tan \left( 4x \right)}}{\dfrac{4{{x}^{2}}\times {{\sin }^{2}}x}{4{{x}^{2}}\times {{\tan }^{2}}\left( 2x \right)}}$
Then, by separating the limits for the functions and by using the formulas stated above to get the answer as:
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{4}\times \dfrac{4x}{\tan \left( 4x \right)}}{\dfrac{1}{4}\dfrac{4{{x}^{2}}}{{{\tan }^{2}}\left( 2x \right)}\dfrac{{{\sin }^{2}}x}{{{x}^{2}}}}$
Now, by solving the above limits by the formulas, we get:
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{4}\times 1}{\dfrac{1}{4}\times 1\times 1}$
Then, we can see that there is no x term left in the question and just by cutting the terms from numerator and denominator, we get:
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{4}\times 1}{\dfrac{1}{4}\times 1\times 1}=1$
So, we get the value of the limit of the function as 1.
Hence, option (d) is correct.

Note:
Now, to solve these type of the questions we can also use the alternate approach as we get the answer by substituting the value of x as 0 in the given function as $\dfrac{0}{0}$, then we can use the L-hospital rule in which we take the differentiation of the numerator and denominator separately and then we substitute the limit to get the final answer.