Question

How do you find the limit of $\dfrac{x\csc x+1}{x\csc x}$as x approaches 0?

Answer 1

To solve this type of problem, first of all we have to simplify the problem. So, we can find the value of the limit easily. As we know $\dfrac{1}{\csc x}=\sin x$ substitute in the given equation. After simplifying the equation apply the identity $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$and then we can find the value of the given limit easily.

Complete step by step answer:
From the question, we are given to find the limit of $\dfrac{x\csc x+1}{x\csc x}$ where x tends to 0.
For solving the question let us consider the above equation as equation (1).
Let us consider
$a=\displaystyle \lim_{x \to 0}\dfrac{x\csc x+1}{x\csc x}.........\left( 1 \right)$
For solving the equation (1) we have to know the limit identity
$\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$
Let us consider the formula as formula ($f_1$)
\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1.........\left( $f_1$ \right)
Let us split the equation (1)
$a=\displaystyle \lim_{x \to 0}\text{ }1+\dfrac{1}{x\csc x}$
Let us consider
$a=\displaystyle \lim_{x \to 0}\text{ }1+\dfrac{1}{x\csc x}.........\left( 2 \right)$
As we know $\dfrac{1}{\csc x}=\sin x$ let us consider this formula as (f2)
$\dfrac{1}{\csc x}=\sin x..........\left( f2 \right)$
Substituting the formula (f2) in equation (2), we get
$a=\displaystyle \lim_{x \to 0}\text{ 1+}\dfrac{\sin x}{x}$
Let us consider the above equation as equation (3).
$a=\displaystyle \lim_{x \to 0}\text{ 1+}\dfrac{\sin x}{x}...........\left( 3 \right)$
By simplifying the equation (3), we get
$a=\text{1+}\displaystyle \lim_{x \to 0}\text{ }\dfrac{\sin x}{x}$
By the formula ($f_1$) we can say that$\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$.
So, therefore let us substitute the value of $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$ in equation (3).
By substituting, we get
$a=\text{1+1}$
$a=2$
So, let us consider the above value as equation (4)
$a=2..........\left( 4 \right)$.

Therefore the limit of $\dfrac{x\csc x+1}{x\csc x}$ is $2$.

Note: To solve this type of problem we have to know all the limit identities perfectly. Because without limit identities we cannot solve the problem. Students should be aware of all trigonometric conversions like $\dfrac{1}{\csc x}=\sin x$. Students should practise more problems to be perfect in this concept.