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Need some help with limits and continuity

  1. Oct 13, 2011 #1
    I have 2 questions in regards to continuity and limits.

    Question 1:
    f(x)= e[itex]^{-x^{2}}[/itex] if x ≠ 0.
    f(x)= c if x=0.

    For which value of c is f(x) continuous at x=0?

    I was thinking the answer would be 1 but I feel that's incorrect.

    Question 2:
    Compute lim x→∞f(x).

    I'm not familiar with how to solve limits to infinity when the variable is in the exponent.
    Any and all help is appreciated! thank you guys.
     
  2. jcsd
  3. Oct 13, 2011 #2

    gb7nash

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    Ok, what does it mean to be continuous at x = 0? Look at your definitions.

    What does the exponent of e approach as x goes to ∞?
     
  4. Oct 13, 2011 #3
    It means that f(x)=f(c) as x→c. So basically that the function can be drawn without having to lift the pen to complete the graph.

    So, e[itex]^{-∞^{2}}[/itex] would be approaching -∞?
     
  5. Oct 13, 2011 #4

    gb7nash

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    Almost. f(x)→f(c) as x→c. In other words:

    [tex]\lim_{x \to c}f(x) = f(c)[/tex]

    So we need to look at the left side and the right side of the equation. Replacing c with 0, what is [itex]\lim_{x \to c}f(x)[/itex]? What is f(c)? Are they equal? If so...


    You're not thinking this all the way through. The exponent is approaching -∞ like you said. What's e raised to a large negative number?
     
  6. Oct 13, 2011 #5
    f(c)=0 and if you were to replace that into the equation for f(x) you would get f(x)=e[itex]^{-0^{2}}[/itex] which would end up equaling 1, right?

    Sorry about this, all of this is really new to me so it's tough to grasp at first.


    Ahh so it would be 0 because e[itex]^{-∞}[/itex] is extremely small.
     
  7. Oct 13, 2011 #6

    gb7nash

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    Correct (I fixed a typo of yours). That's the right side of the equation. Now you need to look at the left side, which is:

    [tex]\lim_{x \to 0}e^{-x^2}[/tex]

    Is this also equal to 1?

    Correct.
     
  8. Oct 13, 2011 #7
    Yes that would be 1.
    So the final answer for making the graph continuous at 0 needs to be 1?
     
  9. Oct 13, 2011 #8

    gb7nash

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    Correct.
     
  10. Oct 13, 2011 #9
    Thank you very much for the help with these questions!!
    Walking me through it helped out a bunch!
     
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