1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Don't forget you also have to observe that f(0)=3.
     
  8. Jul 3, 2012 #7
    Oops, sorry, forgot about that. :redface:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Determining a functions this is continuous at 0
Loading...