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Math Help - Differentiability and continuity

  1. #1
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    Differentiability and continuity

    Hace a problem that asks me to determine if the following function;


    f(x) = \frac{sin(x)}{x} if x<0

    and 1 + x^2 , if x => 0

    is continuous at x = 0 and then if it's differentiable at x=0,

    I figured to compute the limit as x -> 0 from x<0 and x>0 to determine continuity.

    when i did lim(x->0)  \frac{sinx}{x} i used lhopitals to give 1/1 which isnt equal to f(0) = 0 so i figured the function wasn't continuous.
    But i assume i have made a mistake here as when i went ahead and checked for differentiability i found the function to be diff at x=0. which doesn't make sense as if differentiable it must be continuous?

    Please help.

    Cheers.
    Last edited by monster; April 29th 2010 at 09:23 PM.
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  2. #2
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    The function is continuous at 0. The calculus definition of continuity is as followed: A function f is continuous at a point c if both the \displaystyle\lim_{x\to c}f(x) and the f(c) exist and the \lim_{x\to c}f(x)=f(c).
    Last edited by NoobzUseRez; April 29th 2010 at 09:06 PM. Reason: Typo's
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  3. #3
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    Just by looking at the equations you can see they are continuous. The only way \frac{sin(x)}{x} is not continuous is if x can be 0 which you apparently ruled out by limiting the domain to x < 0
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  4. #4
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    sorry, my notations at the start is confusing as i dont know how to do a split function, ill try to fix in original post,

    suppose to be one functions f(x) that equals sinx/x for x<0 and 1+x^2 for x => 0
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  5. #5
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    Quote Originally Posted by NoobzUseRez View Post
    The function is continuous at 0. The calculus definition of continuity is as followed: A function f is continuous at a point c if both the \displaystyle\lim_{x\to c}f(x) and the f(c) exist and the \lim_{x\to c}f(x)=f(c).
    so i just take the limit of the 1+ x^2 part of the function? there is no need to worry about behavior at x<0 ?
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  6. #6
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    If you want to see if a limit exists, show that the limit from the left side equals the limit from the right side. So take the limit of \frac{sin(x)}{x} from the left side, take the limit of x^2 +1 from the right, and see if they equal (which they will).
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  7. #7
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    Quote Originally Posted by monster View Post
    Hace a problem that asks me to determine if the following function;


    f(x) = \frac{sin(x)}{x} if x<0

    and 1 + x^2 , if x => 0

    is continuous at x = 0 and then if it's differentiable at x=0,

    I figured to compute the limit as x -> 0 from x<0 and x>0 to determine continuity.

    when i did lim(x->0)  \frac{sinx}{x} i used lhopitals to give 1/1
    Yes, that is correct.

    which isnt equal to f(0) = 0
    No, f(0)= 1+ 0^2= 1

    so i figured the function wasn't continuous.
    But i assume i have made a mistake here as when i went ahead and checked for differentiability i found the function to be diff at x=0. which doesn't make sense as if differentiable it must be continuous?

    Please help.

    Cheers.
    Follow Math Help Forum on Facebook and Google+

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