Question

How do you find the limit of $\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}$ as x approaches to 0?

Answer 1

We have been given to find the limit of the function whose variable-x is approaching to zero. We shall first substitute $x=0$ in the equation to check the form of the given limit to analyze the method of finding the limit. Then, we will simplify the cosecant and cotangent trigonometric functions in the form of sine and cosine functions to further simplify the function and calculate the limit.

Complete step by step solution:
We have to find $\displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}$.
First, we shall substitute $x=0$ to see whether l’Hospital’s rule can be applied or not.
Putting $x=0$, we get
$\dfrac{0\left( {{\cot }^{2}}0 \right)}{\left( \csc 0 \right)+1}=\dfrac{0}{1}$
This implies that l’Hospitals’s rule cannot be used.
From our basic knowledge of trigonometry, we know that $\csc x=\dfrac{1}{\sin x}$ and $\cot x=\dfrac{\cos x}{\sin x}$. Thus, substituting these values in the given function, we have
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=\displaystyle \lim_{x \to 0}\dfrac{x{{\left( \dfrac{\cos x}{\sin x} \right)}^{2}}}{\left( \dfrac{1}{\sin x} \right)+1}$
$\begin{aligned} & \Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=\displaystyle \lim_{x \to 0}\left[ x\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}.\dfrac{1}{\dfrac{1+\sin x}{\sin x}} \right] \\\ & \Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=\displaystyle \lim_{x \to 0}\left[ x\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}.\dfrac{\sin x}{1+\sin x} \right] \\\ & \Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=\displaystyle \lim_{x \to 0}\left[ \dfrac{x}{\sin x}.\dfrac{{{\cos }^{2}}x}{1+\sin x} \right] \\\ & \Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=\displaystyle \lim_{x \to 0}\left[ \dfrac{1}{\dfrac{\sin x}{x}}.\dfrac{{{\cos }^{2}}x}{1+\sin x} \right] \\\ \end{aligned}$
From the properties of limit, we have $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$. Applying this property, we get
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=\displaystyle \lim_{x \to 0}\left[ \dfrac{1}{1}.\dfrac{{{\cos }^{2}}x}{1+\sin x} \right]$
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=\displaystyle \lim_{x \to 0}\dfrac{{{\cos }^{2}}x}{1+\sin x}$
Finally, we shall put the value of limit in our equation.
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=\dfrac{{{\cos }^{2}}0}{1+\sin 0}$
We also know that $\sin 0=0$ and $\cos 0=1$. Putting these values, we get
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=\dfrac{{{1}^{2}}}{1+0}$
$\Rightarrow \displaystyle \lim_{x \to 0}\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}=1$
Therefore, the value of limit of $\dfrac{x\left( {{\cot }^{2}}x \right)}{\left( \csc x \right)+1}$ as x approaches to 0 is 1.

Note: On applying the value of limit, if the limit comes out to be of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, then we can simply use the l’Hospital’s rule to calculate the limit. According to the l’Hospital’s rule, we shall differentiate the numerator and the denominator separately considering them to be two different functions. Then, we shall keep repeating this procedure until we get a finite value of our limit. Further, we can put the value of the given limit and compute our final answer.