Question

How do you find the $x$ values at which $f\left( x \right)=csc2x$ is not continuous, which of the discontinuities are removable ?

Answer 1

In order to solve the given question, we can solve by considering the definition of discontinuity. We know that a function is said to be discontinuous at ‘a’ then $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ is not equal to$f\left( a \right)$. If the limit does not exist then the function is discontinuous also. In order to find the point of discontinuity of any function, we simply equate the denominator with zero.

Complete step by step answer:
We have given that,$f\left( x \right)=csc2x$. The given f(x) is continuous at a point x = a in domain if;
$\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)$ Exists when it is equal to $f\left( a \right)$.
If the given function is continuous at every point that is in the domain of the function, then we can say that the function $f\left( x \right)$ is continuous and no discontinuities are present in the given function. By the above definition,the function $f\left( x \right)=csc2x$ is continuous everywhere.Using the trigonometric identity,
$\csc 2x=\dfrac{1}{\sin 2x}$
Then the domain of the given function is the set of all the values where sin2x is not equal to 0.So,
$2x\ne \pi \Rightarrow x\ne k\dfrac{\pi }{2}$, where k is any constant term.
So the given function $f\left( x \right)=csc2x$ is discontinuous at $x=k\dfrac{\pi }{2}$.
Thus,
$\underset{x\to {{\left( k\pi \right)}^{+}}}{\mathop{\lim }}\,\csc 2x=\infty \ne -\infty =\underset{x\to {{\left( k\pi \right)}^{-}}}{\mathop{\lim }}\,\csc 2x$
And
$\underset{x\to {{\left( \dfrac{\left( 2k+1 \right)\pi }{2} \right)}^{+}}}{\mathop{\lim }}\,\csc 2x=-\infty \ne \infty =\underset{x\to {{\left( \dfrac{\left( 2k+1 \right)\pi }{2} \right)}^{-}}}{\mathop{\lim }}\,\csc 2x$
Left limit and the right limit disagree at the point $x=k\left( \dfrac{\pi }{2} \right)$.

Therefore, the given function $f\left( x \right)=csc2x$ is discontinuous and non-removable at every value at which sin2x = 0 that is the every multiple of $\dfrac{\pi }{2}$.

Note: Discontinuous function is a function that has a discontinuous graph that means the graph which does not have flow of lines of a given function, then the function is said to be a discontinuous function. A continuous function is a function that creates a continuous graph and has a flow of lines, then the given function is said to be a continuous function.