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Math Help - please help with limits of indeterminate forms

  1. #1
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    please help with limits of indeterminate forms

    Im horrible at this sorry about it being bold .. it wont unbold
    How do you evaluate the following limits

    1) lim x----> 3 [ ( l x-3 l ) / ( x-3 ) ]

    2) lim x --->4/8 [ ( 3x^2 + x - 4 ) / (3x + 4 ) ]

    3) lim x--->1 [ (x^2 + x -2 ) / ( l x -1 l ) ]

    4)lim x--->-1 [ ( 2x^2 +5x + 3 ) / (x+1) ]

    5) lim x---->0 { [ x+8] ^ 1/3 -2} / x
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  2. #2
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    1. \lim_{x\to3}\frac{\lvert x-3\rvert}{x-3}
    Consider the left- and right-sided limits. For x<3, \lvert x-3\rvert=-(x-3) and for x\geq3, \lvert x-3\rvert=x-3. So,

    \lim_{x\to3^-}\frac{\lvert x-3\rvert}{x-3}

    \lim_{x\to3^-}\frac{-(x-3)}{x-3}=-1

    and

    \lim_{x\to3^+}\frac{\lvert x-3\rvert}{x-3}

    \lim_{x\to3^+}\frac{x-3}{x-3}=1.

    What does this tell you about the two-sided limit?

    2. \lim_{x\to4/8}\frac{3x^2+x-4}{3x+4}
    This just requires direct substitution. Did you write the problem correctly?

    3. \lim_{x\to1}\frac{x^2+x-2}{\lvert x-1\rvert}
    \lim_{x\to1}\frac{x^2+x-2}{\lvert x-1\rvert}

    =\lim_{x\to1}\frac{(x-1)(x+2)}{\lvert x-1\rvert}

    Again, consider the one-sided limits.

    4. \lim_{x\to-1}\frac{2x^2+5x+3}{x+1}
    \lim_{x\to-1}\frac{2x^2+5x+3}{x+1}

    =\lim_{x\to-1}\frac{(2x+3)(x+1)}{x+1}

    =\lim_{x\to-1}(2x+3)

    5. \lim_{x\to0}\frac{(x+8)^{1/3}-2}x
    \lim_{x\to0}\frac{(x+8)^{1/3}-2}x=\frac00, so we can use L'H˘pital's rule.

    \lim_{x\to0}\frac{(x+8)^{1/3}-2}x=\lim_{x\to0}\frac13(x+8)^{-2/3}

    =\lim_{x\to0}\frac1{3(x+8)^{2/3}}

    Finish it.
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  3. #3
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    Thank you so much .
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  4. #4
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    For the fifth one just put u^3=x+8.
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