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Math Help - limit problem with radicals

  1. #1
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    limit problem with radicals

    lim [(1+h)^(1/2) - (1-h)^(1/2) - 2] / h
    h~> 0

    "limit as h approaches 0 of [(the square root of 1+h) - (the square root of 1-h) - 2] all over h"

    i plugged in 0 and i get the indeterminate form 0/0 so i tried to multiply by the conjugate. The thing is that i can't seem to get rid of the radical and i can't cancel out the h on the bottom either. no matter what conjugate i try, i can't get rid of the radical and the expression doesn't seem to get any simpler. please help me.

    P.S. i know you can just use L'Hospital's Rule to figure out the limit to be 0, but my teacher wants us to do these problems using algebra and not using that rule.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by oblixps View Post
    lim [(1+h)^(1/2) - (1-h)^(1/2) - 2] / h
    h~> 0

    "limit as h approaches 0 of [(the square root of 1+h) - (the square root of 1-h) - 2] all over h"

    i plugged in 0 and i get the indeterminate form 0/0 so i tried to multiply by the conjugate. The thing is that i can't seem to get rid of the radical and i can't cancel out the h on the bottom either. no matter what conjugate i try, i can't get rid of the radical and the expression doesn't seem to get any simpler. please help me.

    P.S. i know you can just use L'Hospital's Rule to figure out the limit to be 0, but my teacher wants us to do these problems using algebra and not using that rule.
    From the way you typed it out,

    \lim_{h\to0}\frac{\sqrt{1+h}-\sqrt{1-h}-2}{h}=\frac{\sqrt{1}-\sqrt{1}-2}{0}=-\frac{2}{0}. This implies that the limit doesn't exist.

    If you look at right hand and left hand limits, \lim_{h\to0^+}\frac{\sqrt{1+h}-\sqrt{1-h}-2}{h}=-\infty and \lim_{h\to0^-}\frac{\sqrt{1+h}-\sqrt{1-h}-2}{h}=\infty

    Does this make sense?
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  3. #3
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by oblixps View Post
    lim [(1+h)^(1/2) - (1-h)^(1/2) - 2] / h
    h~> 0

    "limit as h approaches 0 of [(the square root of 1+h) - (the square root of 1-h) - 2] all over h"

    i plugged in 0 and i get the indeterminate form 0/0 so i tried to multiply by the conjugate. The thing is that i can't seem to get rid of the radical and i can't cancel out the h on the bottom either. no matter what conjugate i try, i can't get rid of the radical and the expression doesn't seem to get any simpler. please help me.

    P.S. i know you can just use L'Hospital's Rule to figure out the limit to be 0, but my teacher wants us to do these problems using algebra and not using that rule.
    Could you split the numerator into \sqrt{1+h}-1 - (\sqrt{1-h} + 1)?
    Then split it into two fractions, multiply by conjugates in each, and that should solve the problem. Good luck!!

    EDIT: adding some details here.
     \frac{\sqrt{1+h}-1}{h}  - \frac{\sqrt{1-h} + 1}{h}
    Last edited by apcalculus; June 11th 2009 at 04:55 PM.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Finishing apcalcs thought...

    =\lim_{h\to0}\frac{\sqrt{1+h}-1}{h}-\lim_{h\to0}\frac{\sqrt{1-h}-1}{h}

    =\lim_{h\to0}\frac{\sqrt{1+h}-1}{h}\frac{(\sqrt{1+h}+1)}{(\sqrt{1+h}+1)}-\lim_{h\to0}\frac{\sqrt{1-h}+1}{h}\frac{(\sqrt{1-h}-1)}{(\sqrt{1-h}-1)}

    \lim_{h\to0}\frac{(1+h)-1}{h(\sqrt{1+h}+1)}-\lim_{h\to0}\frac{(1-h)-1}{h(\sqrt{1-h}-1)}

    This is not looking good...
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    Finishing apcalcs thought...

    =\lim_{h\to0}\frac{\sqrt{1+h}-1}{h}-\lim_{h\to0}\frac{\sqrt{1-h}-1}{h}

    =\lim_{h\to0}\frac{\sqrt{1+h}-1}{h}\frac{(\sqrt{1+h}+1)}{(\sqrt{1+h}+1)}-\lim_{h\to0}\frac{\sqrt{1-h}+1}{h}\frac{(\sqrt{1-h}-1)}{(\sqrt{1-h}-1)}

    \lim_{h\to0}\frac{(1+h)-1}{h(\sqrt{1+h}+1)}-\lim_{h\to0}\frac{(1-h)-1}{h(\sqrt{1-h}-1)}

    This is not looking good...
    Hence why the limit is undefined...
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  6. #6
    Senior Member apcalculus's Avatar
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    This limit fails to exist indeed... at least we tried

    I think vonnemo19's last algebra step is more plausible than the direct plug in justification. no?
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  7. #7
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    oh no sorry guys i typed the limit wrong. should be "+" instead of "-".

    lim [(1+h)^(1/2) + (1-h)^(1/2) - 2] / h
    h~> 0

    "limit as h approaches 0 of [(the square root of 1+h) + (the square root of 1-h) - 2] all over h"

    yeah so when you plug in 0 for h, you will get the indeterminate form 0/0. my mistake earlier.
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  8. #8
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by oblixps View Post
    lim [(1+h)^(1/2) + (1-h)^(1/2) - 2] / h
    h~> 0

    "limit as h approaches 0 of [(the square root of 1+h) + (the square root of 1-h) - 2] all over h"
    Let f:x\mapsto \sqrt{1+x}+\sqrt{1-x}. f is differentiable at x=0 and note that
    \lim_{h\to0}\frac{\sqrt{1+h}+\sqrt{1-h}-2}{h}=\lim_{h\to0}\frac{f(h)-f(0)}{h}=f'(0).
    Now all you have to do is differentiate f and compute f'(0)\ldots
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