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Math Help - Limit help

  1. #1
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    Limit help

    Any hints to how to solve this type of problem would be great

    f(x) = 6 + ( x^7 / sin(x) ) for x > 0
    f(x) = a for x = 0
    f(x) = 1 / ( x+ 9 ) + b for x < 0

    (peicewise function)

    Find lim x -> 0+ f(x)

    find lim x -> 0- f(x)

    then how would i find what a and b are?

    thanks
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by Stosw View Post
    Any hints to how to solve this type of problem would be great

    f(x) = 6 + ( x^7 / sin(x) ) for x > 0
    f(x) = a for x = 0
    f(x) = 1 / ( x+ 9 ) + b for x < 0
    (peicewise function)
    Find lim x -> 0+ f(x)
    find lim x -> 0- f(x)
    then how would i find what a and b are?
    thanks
    For the limit as x goes to 0 from the positive side you use the first branch of the function, since its domain is x > 0.

    To evaluate the limit, write f(x) = 6 + x^6 \frac{x}{sin(x)}

    and note that \frac{x}{sin(x)} OR \frac{sin(x)}{x} both go to 1 as x approaches 0.

    So the limit becomes 6 + 0* 1 = 6

    For the other limit use the third branch, which gives \frac{1}{9}  + b

    You haven't shared the conditions this problem wants. I assume it's continuity, which means the limits from the left and right are equal to the function value.

    So, two conditions:

    6 = \frac{1}{9} + b (limit exists)

    f(0) = a = 6        (limit equal to function value, so the graph, when you draw it, is filled)

    Good luck!!
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