Why each function is discontinous at given point

• September 22nd 2008, 07:19 AM
Cyberman86
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 (
• September 22nd 2008, 07:37 AM
Shyam
Quote:

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 (

$f\left( x \right) = \left\{ \begin{gathered}
\frac{{x^2 - 1}}
{{x + 1}}{\;\;\;\;\;\text{if }}x \ne - 1 \hfill \\
6{\;\;\;\;\;\;\;\;\;\;\;\; \text{ if }}x = - 1 \hfill \\
\end{gathered} \right.$

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

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

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

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

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

${\text{In the same way, calculate, }}\mathop {\lim }\limits_{x \to - 1 + } f\left( x \right){\text{ and }}f\left( { - 1} \right) \hfill \\$
• September 22nd 2008, 06:44 PM
Krizalid
There's no need to see one sided limits, it suffices to compute $\underset{x\to -1}{\mathop{\lim }}\,\frac{x^{2}-1}{x+1}$ and verify that the value of such limit equals $f(-1).$