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Math Help - i need help with the following limits questions

  1. #1
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    i need help with the following limits questions

    so far i can only do some types but when they vary its confusing!

    lim x^2 − 5x + 2.
    x->4

    lim 5 sin x
    x->0 x

    lim 1
    1 + sin x
    x->0
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  2. #2
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    1. P(x) = x^2 - 5x + 2

    \lim_{x \to a} P(x) = P(a)

    Only for Polynomials!

    2.  f(x) = \frac{5\sin{x}}{x}

    \lim_{x \to 0} f(x) = \frac{0}{0}

    The limit has the indeterminate form of 0/0, so we use L'H˘pital's rule and differentiate the numerator and denominator.

     \lim_{x \to 0} \frac{5\sin{x}}{x}

     \lim_{x \to 0} \frac{5\cos{x}}{1} = 5\cos{0} = 5
    Last edited by Chop Suey; July 8th 2008 at 12:38 PM.
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  3. #3
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    1. P(x) = x^2 - 5x + 2

    Lim P(x) = P(a)
    x -> a

    Only for Polynomials!
    It works for any function continuous at a... after all that is the definition of continuity.


    Quote Originally Posted by Emmett View Post
    lim 1
    1 + sin x
    x->0
    \lim_{x \to 0} \frac1{1+\sin x} = \frac1{1+0} = 1

    If you believe that  \frac1{1+\sin x} is continuous around 0, you can substitute.
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  4. #4
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    Quote Originally Posted by Isomorphism View Post
    It works for any function continuous at a... after all that is the definition of continuity.




    \lim_{x \to 0} \frac1{1+\sin x} = \frac1{1+0} = 1

    If you believe that  \frac1{1+\sin x} is continuous around 0, you can substitute.
    thats really helpful, thanks, just one quick question though, do we take it that sin x is always equal to 0 or was there something else i missed?
    thanks again
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  5. #5
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    When you plug in 0 into the x, and you don't get an indeterminate form, you would get the limit (1).

    Sin(x) equals 0 if x equals 0.
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  6. #6
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    Quote Originally Posted by Chop Suey View Post
    When you plug in 0 into the x, and you don't get an indeterminate form, you would get the limit (1).

    Sin(x) equals 0 if x equals 0.

    oh rite i get it now, thanks dude
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