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Discontinuous/Continuous Functions

  1. Sep 27, 2014 #1
    1. The problem statement, all variables and given/known data

    Is the function continuous at x=1?

    f(x) =
    When x is less than 1:
    [tex](x-1)^2sin(\frac{1}{x-1})[/tex]
    When x is greater or equal to 1:
    [tex]xsin(πx)[/tex]

    2. Relevant equations
    None

    3. The attempt at a solution
    The answer is yes, don't know how to show that, this isn't homework just studying for exams >.<.


    [tex]
    \lim_{x\rightarrow 1+} {xsin(πx)} = 0
    [/tex]
    ----------------------------------
    [tex]
    \lim_{x\rightarrow 1-} {(x-1)^2sin(\frac{1}{x-1})}
    [/tex]

    Here I used:
    [tex]
    \lim_{x\rightarrow 0} {\frac{sin(x)}{x}} = 1
    [/tex]
    But I don't think you can do this, so I don't know how else to solve this.

    [tex]
    \lim_{x\rightarrow 1-} {(x-1)} * \lim_{x\rightarrow 0} {\frac{sin(\frac{1}{x-1})}{\frac {1}{x-1}}} = 0 * 1 = 0
    [/tex]

    Help explaining how to do this question would be helpful.:)
     
  2. jcsd
  3. Sep 27, 2014 #2
    You can't do that, but there is a really simple trick. What are the maximum and minimum values you can have for sin(x)? Are they finite or infinite?
     
  4. Sep 27, 2014 #3
    Sin(x) goes between 1 and -1 and it oscillates to infinity and negative infinity.

    I ended up thinking that as the limit approaches 1 from the left side,
    (x-1)^2 becomes 0 and sin(1/(x-1)) becomes sin(-infinity), so it will be 0 * a number between 1 and -1, and that results in a 0. Is this valid? because sin (-infinity) is undefined, so I don't know if this would be correct or is there a way to show this without the sin being undefined.
     
    Last edited: Sep 27, 2014
  5. Sep 27, 2014 #4
    It's correct. When dealing with limits, a finite number multiplied by zero is still zero, even if you don't know exactly what the finite number is.
     
  6. Sep 27, 2014 #5
    Thanks :)
     
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