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Math Help - Toough Limit Problem

  1. #1
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    Toough Limit Problem

    Hey guys... and gals.. Got one for ya...
    Find (in terms of the constant a)
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  2. #2
    Super Member General's Avatar
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    Quote Originally Posted by desizon View Post
    Hey guys... and gals.. Got one for ya...
    Find (in terms of the constant a)
    \lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h}
    =\lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h} \frac{\sqrt{7(a+h)} + \sqrt{7a}}{\sqrt{7(a+h)} + \sqrt{7a}}

    It should be easy now.
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  3. #3
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    Quote Originally Posted by General View Post
    \lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h}
    =\lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h} \frac{\sqrt{7(a+h)} + \sqrt{7a}}{\sqrt{7(a+h)} + \sqrt{7a}}

    It should be easy now.
    I know up to that part.. This one is still causing problems.
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  4. #4
    Super Member General's Avatar
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    Quote Originally Posted by desizon View Post
    I know up to that part.. This one is still causing problems.
    Surely, you know that If you want to multiply a fraction by another fraction, you will use this formula:
    (\frac{a}{b})(\frac{c}{d})=\frac{ac}{bd}, for b,d \neq 0

    Now, To multiply the numerators in your problem, Use this formula:
    (a-b)(a+b)=a^2-b^2 .
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  5. #5
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    I know that, So I'll have 7(a+h) - 7a on the top, which simplifies down to just

    7h, since the "a" will cancel out.....

    What do I do with the deal on the bottom?
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  6. #6
    Super Member General's Avatar
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    Quote Originally Posted by desizon View Post
    I know that, So I'll have 7(a+h) - 7a on the top, which simplifies down to just

    7h, since the "a" will cancel out.....

    What do I do with the deal on the bottom?
    the bottom is h(\sqrt{7(a+h)} + \sqrt{7a})
    Cancel out the h's.
    and then compute your limit by direct substitution.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by desizon View Post
    Hey guys... and gals.. Got one for ya...
    Find (in terms of the constant a)
    Think derivatives.
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  8. #8
    Super Member Bacterius's Avatar
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    What do I do with the deal on the bottom?
    As General said, you are left with :

    \lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h} \frac{\sqrt{7(a+h)} + \sqrt{7a}}{\sqrt{7(a+h)} + \sqrt{7a}}

    This might look impressive at first, but take the top part of the fraction :

    (\sqrt{7(a+h)} - \sqrt{7a} )(\sqrt{7(a+h)} + \sqrt{7a}).

    This can be simplified to 7(a+h) - 7a = 7(a + h - a) = 7h. So your limit basically boils down to :

    \lim_{h\to0} \frac{7h}{h(\sqrt{7(a+h)} + \sqrt{7a})}

    Cancel the h, and you are pretty much done.
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  9. #9
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    Quote Originally Posted by desizon View Post
    Hey guys... and gals.. Got one for ya...
    Find (in terms of the constant a)
    It has previously been remarked to think derivatives. I will provide a further prompt: Think of differentiating f(x) = \sqrt{7x} from first principles ....
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  10. #10
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    Quote Originally Posted by Bacterius View Post
    As General said, you are left with :

    \lim_{h\to0} \frac{\sqrt{7(a+h)} - \sqrt{7a} }{h} \frac{\sqrt{7(a+h)} + \sqrt{7a}}{\sqrt{7(a+h)} + \sqrt{7a}}

    This might look impressive at first, but take the top part of the fraction :

    (\sqrt{7(a+h)} - \sqrt{7a} )(\sqrt{7(a+h)} + \sqrt{7a}).

    This can be simplified to 7(a+h) - 7a = 7(a + h - a) = 7h. So your limit basically boils down to :

    \lim_{h\to0} \frac{7h}{h(\sqrt{7(a+h)} + \sqrt{7a})}

    Cancel the h, and you are pretty much done.
    Thanks... I'm still not getting the right answer, so, what I have, is 6h left on top, and the bottom is?
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  11. #11
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    Quote Originally Posted by desizon View Post
    Thanks... I'm still not getting the right answer, so, what I have, is 6h left on top, and the bottom is?
    how do you get 6h?

    \frac{7h}{h\left[\sqrt{7(a+h)}+\sqrt{7a}\right]} = \frac{7}{\sqrt{7(a+h)}+\sqrt{7a}}<br />

    now let h \to 0
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  12. #12
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    edit:::
    Thanks all. Got it.
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