Question

Find the points of discontinuity of the function f, where ${\displaystyle f\left(x\right)=\{\begin{array}{lll}x+2\text{,}& \text{if}& x\ge 2\\ x^2\text{,}& \text{if}& x<2\end{array}}$

Answer 1

${\displaystyle \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}{x}^2}$

${\displaystyle \underset{x\to 2^{-}}{lim}f\left(x\right)}$ = 2² = 4  .......(1)

${\displaystyle \underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(x+2)}$

${\displaystyle \underset{x\to 2^{+}}{lim}f\left(x\right)}$ = 2 + 2 = 4  .......(2)

From equation (1) and (2), we get

${\displaystyle \underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)}$

∴ ${\displaystyle \underset{x\to 2}{lim}f\left(x\right)}$ exist.

Let x₀ be an arbitrary point such that x₀ < 2.

Then ${\displaystyle \underset{x\to x_0}{lim}f\left(x\right)= \underset{x\to x_0}{lim}{x}^2}$

${\displaystyle \underset{x\to x_0}{lim}f\left(x\right)= x_0^2}$

${\displaystyle f\left(x_0\right)=x_0^2}$

∴ ${\displaystyle \underset{x\to x_0}{lim}f\left(x\right)=f\left(x_0\right)}$

For the point x₀ < 2, we have the limit of the function that exists and is equal to the value of the function at that point.

Since x₀ is an arbitrary point the above result is true for all x < 2.

∴ f(x) is continuous in ${\displaystyle (-\infty ,2)}$.

Let x₀ be an arbitrary point such that x₀ > 2

Then ${\displaystyle \underset{x\to x_0}{lim}f\left(x\right)= \underset{x\to x_0}{lim}(x+2)}$

= x₀ + 2

${\displaystyle f\left(x_0\right)=x_0+2}$

∴ ${\displaystyle \underset{x\to x_0}{lim}f\left(x\right)=f\left(x_0\right)}$

∴ For the point x₀ > 2, the limit of the function exists and is equal to the value of the function.

Since x₀ is an arbitrary point the above result is true for all x > 2.

∴ The function is continuous at all points of ${\displaystyle (2,\infty )}$.

Hence the given function is continuous at all points of R.