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Math Help - Limits - Solve By Manipulation

  1. #1
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    Limits - Solve By Manipulation

    Okay, so I know you have to rationalize the limit in order to get it out of the indeterminate form, but I can't seem to do it for these two questions. Can someone help, start me off?

    a) Lim as x -> 4 of (x^2-4)/(sqrtx - 2)

    b) Lim as x -> 0 of [sqrt(3-x) - sqrt(x+3)] / x

    Thanks,
    Stefan
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  2. #2
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    a) \lim_{x \to 4} \frac{x^2 - 4}{\sqrt{2}-2} = \left[\frac{12}{0}\right]^+

    This is not an indeterminate form. You'll find that the limit does not exist. Consider the left and right handed limits and you'll see they're not equal.

    _____________

    b) \begin{aligned} \lim_{x \to 0} \frac{\sqrt{3-x} - \sqrt{x+3}}{x} \cdot {\color{blue} \frac{\sqrt{3-x} + \sqrt{x+3}}{\sqrt{3-x} + \sqrt{x+3}}}\\   = \lim_{x \to 0} \frac{3-x - (x+3)}{x\left(\sqrt{3-x}+\sqrt{x+3}\right)} \\  = \lim_{x \to 0} \frac{-2x}{x\left(\sqrt{3-x}+\sqrt{x+3}\right)} \\ \end{aligned}

    Notice that something cancels Then you can plug in x = 0 without getting an indeterminate form.
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  3. #3
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    Hi, thanks or the help, a quick question though. When we rationalize why do we use the opposite sign, for example, (b) had sqrt + sqrt and when it was rationalized it was sqrt - sqrt. What is the actual reason this is done, I know it cancels but what proof is there that that always works?
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  4. #4
    o_O
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    Well, it all boils down to the difference of squares formula: (a-b)(a+b) = a^2 - b^2

    In particular, if a and b are square roots, we would like to get rid of them and by multiplying by the conjugate (in this case, (a+b)) we get rid of them and we get an expression without square roots, i.e. a^2 - b^2
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