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Math Help - Limits

  1. #1
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    Smile Limits

    These are typical examples of the limits I am having trouble on. I need help and I have test on Monday eek I understand most limits I just have trouble with these. =D Thank You-- Carol-ANN

    PS. I know how to do number 4
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  2. #2
    Senior Member DeMath's Avatar
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    For 1)

    \mathop {\lim }\limits_{\Delta x \to 0}  - \frac{{\sqrt {\left( {x + \Delta x} \right) - 8}  - \sqrt {x + 8} }}{{\Delta x}} =

    =  - \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left( {\sqrt {\left( {x + \Delta x} \right) - 8}  - \sqrt {x - 8} } \right)\left( {\sqrt {\left( {x + \Delta x} \right) - 8}  + \sqrt {x - 8} } \right)}}<br />
{{\Delta x\left( {\sqrt {\left( {x + \Delta x} \right) - 8}  + \sqrt {x - 8} } \right)}} =

    =  - \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\left( {x + \Delta x} \right) - 8 - \left( {x - 8} \right)}}<br />
{{\Delta x\left( {\sqrt {\left( {x + \Delta x} \right) - 8}  + \sqrt {x - 8} } \right)}} =

    =  - \mathop {\lim }\limits_{\Delta x \to 0} \frac{1}{{\sqrt {\left( {x + \Delta x} \right) - 8}  + \sqrt {x - 8} }} =  - \frac{1}{{2\sqrt {x - 8} }}.
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  3. #3
    Member eXist's Avatar
    Joined
    Aug 2009
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    For #2)
    Substitute and apply Hospital's rule:

    \lim_{x\rightarrow0}\frac{1-cos^2x}{x} = <br />
\lim_{x\rightarrow0}\frac{sin^2x}{x} =<br />
\lim_{x\rightarrow0}\frac{\frac{d}{dx}(sin^2x)}{\f  rac{d}{dx}(x)} =<br />
\lim_{x\rightarrow0}\frac{2sinxcosx}{1} =<br />
    <br />
\frac{2sin(0)cos(0)}{1}<br />
= 0<br />

    For #3)
    Look at this graph:

    As x\rightarrow\frac{\pi^+}{2} you get -\infty

    For #5)
    It's a similar problem. You're approaching form the right (positive side) so substitute 1.1 in. You're going to get a really large negative number. Now imagine as 1.1\rightarrow1.0 it is going to approach -\infty
    Last edited by eXist; August 30th 2009 at 01:46 PM.
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