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Limit does not exist but function exist

  1. Jun 19, 2015 #1
    1. The problem statement, all variables and given/known data
    there are four cases on limits given to us, and one of them I didnt really understand.
    This case was: Limit f(x) as x approaches a does not exist but f(a) exist.
    2. Relevant equations
    upload_2015-6-20_8-32-29.png
    3. The attempt at a solution
    My answer here that the limit in this piecewise defined function exist,
    As x approaches -5 from the left, it uses the function x+3;
    And as x approaches -5 from the right, it uses the function √(25-x^2)


    please give me a great explanation with this case

    Thank you
     
  2. jcsd
  3. Jun 19, 2015 #2

    Ray Vickson

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    Do you really think that 0 = -2? That would need to be true in order that f(x) have a limit as x → -5.
     
  4. Jun 19, 2015 #3

    Fredrik

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    This suggests that the limit should be -2.

    This suggests that the limit should be 0.

    A limit of f at 5 is a number L such that for all ##\varepsilon>0## there's a ##\delta>0## such that the following implication holds for all real numbers ##x##.
    $$0<|x-5|<\delta\ \Rightarrow\ |f(x)-L|<\varepsilon.$$ The implication is saying that f maps the interval ##(5-\delta,5+\delta)## into the interval ##(L+\varepsilon,L-\varepsilon)##.

    Now let's see if there a ##\delta>0## with the property above, in the special case when ##\varepsilon=1##. No matter how small we choose ##\delta##, there's an x in the interval ##(5-\delta,5+\delta)## such that f(x) is not in the interval (-2-1,-2+1)=(-3,-1). (Choose x slightly greater than 5, so that f(x) is very close to 0). This implies that -2 can't be a limit. And there's also a y in the interval ##(5-\delta,5+\delta)## such that f(y) is not in the interval (0-1,0+1)=(-1,1). (Choose y slightly less than 5, so that f(y) is very slose to -2). This implies that 0 can't be a limit.
     
    Last edited: Jun 19, 2015
  5. Jun 19, 2015 #4

    HallsofIvy

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    Yes, and so as x gets closer and closer to -5 (from the left), f(x) gets closer and closer to -5+ 3= -2.

    Yes, and so as x gets closer and closer to -5 (from the right), f(x) gets closer and closer to √(25-(-5)^2)= 0.
    But we can't say that f(x) gets closer and closer to any one number as x gets closer and closer to -5.

     
  6. Jun 22, 2015 #5
    thank you for the great explanation
     
  7. Jun 23, 2015 #6

    SammyS

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    There is a typo on that page the image is taken from:

    upload_2015-6-23_12-0-36.png

    It seems this should have said:

    ##\displaystyle\ \lim_{x\to -5}\, f(x) \ \text{ does not exist. [why?]} \ ##​

    As it's stated in your image, the limit does exist. It's simply the same as the constant ƒ(-5) , which is zero .
     
  8. Jul 4, 2015 #7
    Does the problem say "there exists an a such that..." or "for all a.."

    Also, are you required to give an equation as the representation of a function? A function can be represented in 4 different ways: in words, as an equation, in a table, and as a graph. Perhaps reasoning with one of these other representations may be helpful.
     
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