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Math Help - Determining limits.

  1. #1
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    Determining limits.

    just want to check something.

    Determine limits:

    lim x -> -5 (x +4)^2012

    would the answer be one?

    But i got that answer just by looking at the answer and subtracting the 4 from the negative 5 and i dont think that's right.

    Whats the answer and how does the working look?

    Thanks.
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  2. #2
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    Re: Determining limits.

    Quote Originally Posted by Hooperoo View Post
    just want to check something.

    Determine limits:

    lim x -> -5 (x +4)^2012

    would the answer be one?

    But i got that answer just by looking at the answer and subtracting the 4 from the negative 5 and i dont think that's right.

    Whats the answer and how does the working look?

    Thanks.
    The function is continuous at the point x = -5, so the limit of the function is equal to the function evaluated at x = -5.
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  3. #3
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    Re: Determining limits.

    Any chance you can help me on this one mate?

    lim x -> 4 (4-x)/(5-sqrtx^2 + 9)

    i subbed in x = 4 and solved and got 0. right?
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  4. #4
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    Re: Determining limits.

    it's hard to tell exactly what you mean. The most direct interpretation would be
    4\frac{4- x}{5- \sqrt{x}+ 9}
    At x= 4, that the numerator is 0 while the denominator is 5- 4+ 9= 10 so the limit is 0/10= 0.

    But you might also mean
    4\frac{4- x}{5- \sqrt{x+ 9}}
    At x= 4, the numerator is still 0 while the denominator is [itex]5- \sqrt{13}[/itex][ which is NOT 0 so the limit is still 0.

    These are pretty trivial specifically because the denominator is not 0. At basic property of limits is
    "If [tex]\lim_{x\to a} f(x)= A[/itex] and [itex]\lim_{x\to a} g(x)= B\ne 0[/itex], then
    \lim_{x\to a}\frac{f(x)}{g(x)}= \frac{A}{B}"

    If the denominator does go to 0, then we have:
    1) if the numerator does not go to 0, the limit does not exist.
    2) if both numerator and denominator go to 0, the limit may or may not exist. We need to try to "reduce" the fraction to eliminate this possiblitiy. In particular, if both numerator and denominator are polynomials that are 0 at x= a, they both have a factor of x- a which we can then cancel.
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