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Homework Help: Determining a functions this is continuous at 0

  1. Jul 3, 2012 #1
    1. The problem statement, all variables and given/known data

    f(x) = {2x2 + x +3, x < 0
    [itex]\frac{3}{x + 1}[/itex] x ≥ 0

    The 2 should be wrapped as 1 with a { but do not know how to do that.



    2. Relevant equations



    3. The attempt at a solution

    I was wondering if the squeeze rule would be relevant here? but I do not know how to derive g(x) and h(x) from the above so that I can show g(x) ≤ f(x) ≤ h(x)

    Once I know that then I think I can continue. I can already see that if f(0) then both would = 3
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 3, 2012 #2

    Simon Bridge

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    Here, let me:[tex]f(x)=\left \{
    \begin{array}{rl}
    2x^2+x+3, & x<0\\ \\ \frac{3}{x+1}, & x \geq 0

    \end{array}\right .[/tex]

    Why not just check that the left and right hand limits are the same?
    http://www.sparknotes.com/math/calcab/functionslimitscontinuity/section3.rhtml
     
    Last edited: Jul 3, 2012
  4. Jul 3, 2012 #3

    SammyS

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    No need for the squeeze theorem.

    Find the limit as x approaches zero from the right, then from the left. Are those limits equal? Do they equal f(0) ?

    One of many ways to display the function in LaTeX:

    [itex]\displaystyle f(x)=\left\{\matrix{2x^2+x+3,\ \ x<0\\ \\ \frac{3}{x+1},\ \ x\ge0 }\right.[/itex]

    The LaTeX code:
    Code (Text):
    [itex]\displaystyle f(x)=\left\{\matrix{2x^2+x+3,\ \ x<0
    \\ \\
     \frac{3}{x+1},\ \ x\ge0 }\right.[/itex]
     
  5. Jul 3, 2012 #4
    All I have written then is this

    xn → 0 [itex]\Rightarrow[/itex] 2xn+xn+3 → 3
    where xn < 0

    and

    xn→ 0 [itex]\Rightarrow[/itex] [itex]\frac{3}{x+1}[/itex] → 3
    where xn≥0

    This holds by the combination rules for sequences therefore f is continuous at 0.

    Is this all I have to write as it doesn't seem to be enough??
     
  6. Jul 3, 2012 #5
    I don't see any xn in the question. :confused:

    Well that seems correct if you mean this:
    [tex]\lim_{x→0^-} 2x^2+x+3=3[/tex]
    [tex]\lim_{x→0^+} \frac{3}{x+1}=3[/tex]

    Yes, its enough to show that if the left and right hand limits are equal, the function is continuous.
     
  7. Jul 3, 2012 #6

    LCKurtz

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    Don't forget you also have to observe that f(0)=3.
     
  8. Jul 3, 2012 #7
    Oops, sorry, forgot about that. :redface:
     
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