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Last updated date: 26th Nov 2023
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Evaluate the following limit: $\displaystyle \lim_{x \to 0}\left( x\csc x \right)$

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Hint: To solve this question we need to have the knowledge of limits. To solve the problem where the function given is $x\csc x$, we will first convert the trigonometric function$\csc x$ in terms of $\sin x$ and then we will be applying the value of the limit which is given to us as $x$ tends to zero.

The question asks us to find the value of the function$\displaystyle \lim_{x \to 0}\left( x\csc x \right)$. We are given with a function which is a product of the algebraic function $x$ and the trigonometric function$\csc x$. The first step is to change the trigonometric function $\csc x$ into the terms of $\sin x$. So as you know, the trigonometric function $\csc x$ is the reciprocal of the trigonometric function $\sin x$. Mathematically it will be written as:
$\Rightarrow \csc x=\dfrac{1}{\sin x}$
$\Rightarrow x\csc x=x\dfrac{1}{\sin x}$
$\Rightarrow x\csc x=\dfrac{x}{\sin x}$
Now to find the value of $\displaystyle \lim_{x \to 0}\left( x\csc x \right)$which has become as $\displaystyle \lim_{x \to 0}\left( \dfrac{x}{\sin x} \right)$ we will divide both the numerator and the denominator of the function by $x$ , on doing this we get:
$\Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{\dfrac{x}{x}}{\dfrac{\sin x}{x}} \right)$
On checking the numerator of the function we see that it cancels to give the result as $1$ as $\dfrac{x}{x}$ is $1$. Now on applying the limit to the function $\dfrac{\sin x}{x}$. As we know that $\displaystyle \lim_{x \to 0}\left( \dfrac{\sin x}{x} \right)=0$, so applying this in the above expression we get:
$\Rightarrow \displaystyle \lim_{x \to 0}\left( \dfrac{1}{\dfrac{\sin x}{x}} \right)$
$\Rightarrow \dfrac{1}{1}$
Since any fraction having denominator as $1$, then the numerator of the fraction changes to an integer. So on doing this we get:
$\Rightarrow 1$
$\therefore$ The value of $\displaystyle \lim_{x \to 0}\left( x\csc x \right)$ is $1$.

Note: To solve the problem like this we need to first solve the function and then need to substitute the value of the limit to the function. When the fraction has a denominator as $1$, then the numerator of the fraction changes to an integer.