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Math Help - is this function contineous

  1. #1
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    is this function contineous

    i have a question.
    is a function limit x tending to 1 f(x)= (x^2-1)/(x-1)
    is contineous at every point or it is discontineous at x=1?

    also the simplified form of above function i.e (x+1) has the same contineuity as the above function??
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by pravinjaggu View Post
    i have a question.
    is a function limit x tending to 1 f(x)= (x^2-1)/(x-1)
    is contineous at every point or it is discontineous at x=1?

    also the simplified form of above function i.e (x+1) has the same contineuity as the above function??
    \lim_{x\to1}\frac{x^2-1}{x-1}=\lim_{x\to1}\frac{\left(x+1\right)\left(x-1\right)}{x-1}=\lim_{x\to1}x+1=2.

    However, f\left(x\right)=\frac{x^2-1}{x-1} is discontinuous at x=1. There is a hole in the function at that value since x-1 is a term common to the numerator and denominator of the function.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by pravinjaggu View Post
    i have a question.
    is a function limit x tending to 1 f(x)= (x^2-1)/(x-1)
    is contineous at every point or it is discontineous at x=1?

    also the simplified form of above function i.e (x+1) has the same contineuity as the above function??
    f(x) is discontinuos at x=1 because it is not DEFINED there.

    This is a REMOVABLE discontinuity because

    f(x)=x+1\forall{x\neq1}
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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    f(x) is discontinuos at x=1 because it is not DEFINED there.

    This is a REMOVABLE discontinuity because

    f(x)=x+1\forall{x\neq1}
    Hello there is the graphe
    Attached Thumbnails Attached Thumbnails is this function contineous-21.jpg  
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