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Math Help - Continuity

  1. #1
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    Continuity

    f(x) = { x ; x<= 0
    x^2 ; 0 < x < =1
    2/x ; 1 < x <= 2
    x-1 ; x>2

    My problem is whether f(x) is continuous at x = 0 or not. Because left hand limit = f(0) when x approaches to 0 but the right hand limit \neq f(0) when x approaches to 0. is it continuous ?
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  2. #2
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    The left hand limit is 0, since the function is f(x) = x to the left of x = 0.
    The right hand limit is 0, since the function is f(x) = x^2 to the right of x = 0.

    Since the left hand limit and right hand limit are equal, the function is continuous at x = 0.
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  3. #3
    Ted
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    @ Reply #2 :

    Limit from right at x=a = Limit from left x=a does not mean the function continuous at a

    this means the limit at x=a exist

    For the continuity, Limit at x=a must equal f(a)
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    Quote Originally Posted by Prove It View Post
    The left hand limit is 0, since the function is f(x) = x to the left of x = 0.
    The right hand limit is 0, since the function is f(x) = x^2 to the right of x = 0.

    Since the left hand limit and right hand limit are equal, the function is continuous at x = 0.
    you mean right hand limit = f(0) = 0 ? even when f(x) = x^2 when 0 < x <=1 ? i thought it is not equal to f(0) when x is not equal to 0.
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    Quote Originally Posted by wolf12 View Post
    you mean right hand limit = f(0) = 0 ? even when f(x) = x^2 when 0 < x <=1 ? i thought it is not equal to f(0) when x is not equal to 0.
    When evaluating limits, you aren't concerned with what the function is AT a point, only what happens to a function NEAR a point.

    And Ted, that is a fine yet true distinction, but since the limit DOES equal f(0), the function is continuous.
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    Quote Originally Posted by Prove It View Post
    When evaluating limits, you aren't concerned with what the function is AT a point, only what happens to a function NEAR a point.

    And Ted, that is a fine yet true distinction, but since the limit DOES equal f(0), the function is continuous.
    f(x) = { x ; x<= 0
    x^2 ; 0 < x < =1
    2/x ; 1 < x <= 2
    x-1 ; x>2

    if the first section (x ; x<= 0) is x ; x< 0, is it still continuous ?
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  7. #7
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    I suggest you draw the graph... A picture tells a thousand words...
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    Quote Originally Posted by Prove It View Post
    I suggest you draw the graph... A picture tells a thousand words...
    i drew, yeah last question was stupid :P, ok i was confused about x ; x<= 0 , x^2 ; 0 < x < =1, now i get it, if you can equal x to 0 from one side, it doesn't matter even if you cant equal x to 0 from the other side right ?
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  9. #9
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    The point is that the function needs to approach the same value from both sides. If you follow the function from the left, do you get to the same point as you would if you followed the function from the right?
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  10. #10
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    Quote Originally Posted by Prove It View Post
    The point is that the function needs to approach the same value from both sides. If you follow the function from the left, do you get to the same point as you would if you followed the function from the right?
    I dont understand that, when i approach from left side at x = 0 f(x) = x = 0, but when i approach from the right side, i can say at x = 0.1, f(x) = x^2 = 0.01, but i cant say its 0, its just getting near 0. Do u take the function as continuous in this situation?
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  11. #11
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    Quote Originally Posted by wolf12 View Post
    I dont understand that, when i approach from left side at x = 0 f(x) = x = 0, but when i approach from the right side, i can say at x = 0.1, f(x) = x^2 = 0.01, but i cant say its 0, its just getting near 0. Do u take the function as continuous in this situation?
    Sort of. Like I said, when dealing with limits, you do not worry about what the function equals at a point. All you want to know is what the function tends to NEAR that point.

    I think you can see that as you get closer and closer to x = 0 on each side, you get closer and closer to f(x) = 0... Since this happens on both sides, the limit exists, and since the function is defined at that point and is equal to the limit, the function is continuous. Does that make sense?
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