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Math Help - piecewise limits

  1. #1
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    piecewise limits

    1.
    Let y={x+1 if x>=1
    {-x^2 if x<1
    Does lim x->1 f(x) exists?

    2.
    y={ax+1 if x>=1
    {-x^2 if x<1
    Find all values of a so that lin x->1 f(x) exists
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bobby77 View Post
    1.
    Let y={x+1 if x>=1
    {-x^2 if x<1
    Does lim x->1 f(x) exists?
    For this limit to exist requires that

    lim(x->1+) f(x) = lim(x->1-) f(x)

    and both be finite (lim(x->1+) means the limit as x goes to 1 from above,
    and lim(x->1-) means the limit as x goes to 1 from below.]

    Now:

    lim(x->1+) f(x) = lim(x->1+) x+1 = 2

    and

    lim(x->1-) f(x) = lim(x->1-) -x^2 = -1.

    Hence:

    lim(x->1+) f(x) != lim(x->1-) f(x)

    and so the limit does not exist.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by bobby77 View Post
    1.
    2.
    y={ax+1 if x>=1
    {-x^2 if x<1
    Find all values of a so that lin x->1 f(x) exists
    As before for the limit to exist requires that

    lim(x->1+) f(x) = lim(x->1-) f(x)

    and both be finite (lim(x->1+) means the limit as x goes to 1 from above,
    and lim(x->1-) means the limit as x goes to 1 from below.]

    Now:

    lim(x->1+) f(x) = lim(x->1+) ax+1 = a+1

    and

    lim(x->1-) f(x) = lim(x->1-) -x^2 = -1.

    Hence:

    lim(x->1+) f(x) = lim(x->1-) f(x)

    only if a+1=-1, which is equivalent to a=-2.

    RonL
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