Question

How do you use the epsilon-delta definition of continuity to prove $f\left( x \right)={{x}^{2}}$ is continuous?

Answer 1

Hint: To solve this question, we use the epsilon delta function given by the general definition:
$\displaystyle \lim_{x \to a}f\left( x \right)=L$ if $\forall \varepsilon >0,\exists \delta >0:$
$\Rightarrow 0<\left| x-a \right|<\delta $
$\Rightarrow \left| f\left( x \right)-L \right|<\varepsilon $
We use this concept and show that the equation $f\left( x \right)={{x}^{2}}$ is continuous.

Complete step by step solution:
The given question requires us to use the epsilon delta function which is a property of limits used to solve questions involving limits. The general form of the epsilon delta function is given by,
$\displaystyle \lim_{x \to a}f\left( x \right)=L$ if $\forall \varepsilon >0,\exists \delta >0:$
$\Rightarrow 0<\left| x-a \right|<\delta $
$\Rightarrow \left| f\left( x \right)-L \right|<\varepsilon $
According to this above definition, it says that if we have a function whose limit tending to a, is a value L for all $\varepsilon <0$ there exists $\delta >0,$ such that the difference between the function variable and the limit value is less than $\delta .$ This implies that the magnitude of the difference between the function $f\left( x \right)$ and the limit value L is less than $\varepsilon .$
Now we have $f\left( x \right)$ as continuous for $x=\overline{x},$ for any real number $\varepsilon >0,$ we can find a real number $\delta >0$ such that x lies in the interval,
$\Rightarrow \left( \overline{x}-\delta \right)This above equation implies that,
$\Rightarrow \left| f\left( x \right)-f\left( \overline{x} \right) \right|<\varepsilon \ldots \ldots \left( 1 \right)$
Here, we have considered the value of the limit L to be $f\left( \overline{x} \right)$ and the limit a as $\overline{x}$ .
Now for the given interval of x,
$\Rightarrow \left| x-\overline{x} \right|<\delta $
$\Rightarrow \left| x \right|<\left| \overline{x} \right|+\delta \ldots \ldots \left( 2 \right)$
Now we consider equation 1,
$\Rightarrow \left| f\left( x \right)-f\left( \overline{x} \right) \right|=\left| {{x}^{2}}-{{\overline{x}}^{2}} \right|$
We can split the two as ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right).$
$\Rightarrow \left| f\left( x \right)-f\left( \overline{x} \right) \right|=\left| \left( x-\overline{x} \right)\left( x+\overline{x} \right) \right|$
This can be individually split as their respective magnitudes,
$\Rightarrow \left| f\left( x \right)-f\left( \overline{x} \right) \right|=\left| x-\overline{x} \right|\left| x+\overline{x} \right|$
This value is $\left| x-\overline{x} \right|$ is nothing but $\delta .$
$\Rightarrow \left| f\left( x \right)-f\left( \overline{x} \right) \right|=\left| x-\overline{x} \right|\left| x+\overline{x} \right|<\delta \left( \left| x \right|+\left| \overline{x} \right| \right)$
Substituting the value of x from equation 2,
$\Rightarrow \left| f\left( x \right)-f\left( \overline{x} \right) \right|<\delta \left( 2\left| \overline{x} \right|+\delta \right)$
Hence, for any value of $\varepsilon >0,$
$\Rightarrow \delta \left( 2\left| \overline{x} \right|+\delta \right)<\varepsilon $
Hence, the condition of continuity is proved and we have used the epsilon-delta definition of continuity to prove $f\left( x \right)={{x}^{2}}$ is continuous.

Note: It is essential to know how to solve for the continuity of the equation by using this special case of epsilon delta method. It is very useful to determine the continuity of functions without actually going into the very depth of the equation. We need to note that this epsilon delta criterion holds true if and only if the graph of the given function has no discontinuities. It works only for continuous functions.