Thread: Why each function is discontinous at given point

1. Why each function is discontinous at given point

How can I explain each function is discontinous at given point.

f(x) x^2 - 1 / over x + 1

at x = -1

--------------------------------------------------------------------

________( x^2 -1 / over x +1, if x doesn't equal -1
F(x)= ___(
________( 6, if x = -1

It should be a big parenthesis but i couldn't draw it here, so i put three (

2. Originally Posted by Cyberman86
How can I explain each function is discontinous at given point.

f(x) x^2 - 1 / over x + 1

at x = -1

--------------------------------------------------------------------

________( x^2 -1 / over x +1, if x doesn't equal -1
F(x)= ___(
________( 6, if x = -1

It should be a big parenthesis but i couldn't draw it here, so i put three (
$\displaystyle f\left( x \right) = \left\{ \begin{gathered} \frac{{x^2 - 1}} {{x + 1}}{\;\;\;\;\;\text{if }}x \ne - 1 \hfill \\ 6{\;\;\;\;\;\;\;\;\;\;\;\; \text{ if }}x = - 1 \hfill \\ \end{gathered} \right.$

$\displaystyle {\text{A function is continuous at a given point }}x = a{\text{ when, }} \hfill \\$

$\displaystyle {\text{Left Hand Limit = Right Hand Limit = }}f\left( a \right) \hfill \\$

$\displaystyle \Rightarrow \mathop {\lim }\limits_{x \to a - } f\left( x \right) = \mathop {\lim }\limits_{x \to a + } f\left( x \right) = f\left( a \right) \hfill \\$

$\displaystyle \Rightarrow \mathop {\lim }\limits_{x \to - 1 - } f\left( x \right) = \mathop {\lim }\limits_{x \to - 1 + } f\left( x \right) = f\left( { - 1} \right) \hfill \\ \hfill \\$

$\displaystyle {\text{Now,}} \hfill \\$
$\displaystyle \mathop {\lim }\limits_{x \to - 1 - } f\left( x \right) = \mathop {\lim }\limits_{x \to - 1 - } \frac{{x^2 - 1}} {{x + 1}} = \mathop {\lim }\limits_{x \to - 1 - } \frac{{\left( {x + 1} \right)\left( {x - 1} \right)}} {{x + 1}} = \mathop {\lim }\limits_{x \to - 1 - } \left( {x - 1} \right) = - 2 \hfill \\$

$\displaystyle {\text{In the same way, calculate, }}\mathop {\lim }\limits_{x \to - 1 + } f\left( x \right){\text{ and }}f\left( { - 1} \right) \hfill \\$

3. There's no need to see one sided limits, it suffices to compute $\displaystyle \underset{x\to -1}{\mathop{\lim }}\,\frac{x^{2}-1}{x+1}$ and verify that the value of such limit equals $\displaystyle f(-1).$