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Thread: Why each function is discontinous at given point

  1. #1
    Junior Member Cyberman86's Avatar
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    Question 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 (
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  2. #2
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    Quote Originally Posted by Cyberman86 View Post
    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 \\ $
    Last edited by Shyam; Sep 22nd 2008 at 07:04 AM.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    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 $\displaystyle f(-1).$
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