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Math Help - limit problem

  1. #1
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    limit problem

    <br />
=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x}<br />

    could you please help me to solve this problem please. Thank you.
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  2. #2
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    Quote Originally Posted by thangbe View Post
    <br />
=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x}

    could you please help me to solve this problem please. Thank you.
    Several approaches are possible.

    There's a simple one if you know differential calculus ..... you can consider the function f(a) = \frac{1}{\sqrt{a + 2}}. Then, with a (perhaps disquieting) change from the usual notation ( x instead of h, a instead of x) you can consider the derivative of f(a) from first principles:

    f'(a) = \lim_{x \rightarrow 0} \frac{f(a + x) - f(a)}{x}.

    So the answer will be the derivative of f(a) with respect to a ....
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  3. #3
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    Quote Originally Posted by thangbe View Post
    <br />
=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x}<br />

    could you please help me to solve this problem please. Thank you.
    If you set x= 0, the numerator is \frac{1}{\sqrt{a+ 2}}- \frac{1}{\sqrt{a}} which is clearly NOT equal to 0 but the denominator is 0. Obviously this limit does not exist.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    If you set x= 0, the numerator is \frac{1}{\sqrt{a+ 2}}- \frac{1}{\sqrt{a}} which is clearly NOT equal to 0 but the denominator is 0. Obviously this limit does not exist.
    Dang it. I misread the red x for a 2:
    ] <br />
=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+{\color{red}x}}}}{x}<br />
    Hmmmmmm ... I bet that red x is meant to be a 2 ....
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  5. #5
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    the limit is exists and the answer is.
    <br /> <br />
=\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x} =\frac {-1}{2{(a+2)}^\frac{3}{2}} <br /> <br />



    but I don't know how to show for it.

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  6. #6
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    Quote Originally Posted by thangbe View Post
    the limit is exists and the answer is.

    =\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x} =\frac {-1}{2{(a+2)}^\frac{3}{2}}

    " alt="

    =\lim _{x->0} \frac{\frac{1}{\sqrt{a+x+2}} - \frac{1}{\sqrt{a+x}}}{x} =\frac {-1}{2{(a+2)}^\frac{3}{2}}

    " />



    but I don't know how to show for it.

    So you have mistyped the question as I suspected - see post #4.

    One way of getting the answer is given in post #2. Study it. If you do not understand that method, then you will need to say so. There are several other methods that can be used.
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  7. #7
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    I understand in your post #2 that why I get that answer. However my teacher didn't allows us to use that menthod thank you for you help sir.
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  8. #8
    MHF Contributor matheagle's Avatar
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    Clearly there is a typo.
    But once that is resolved, my favourite way to solve these type of limits is by multiplying by the conjugate. Not many people teach that these days.
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  9. #9
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    Quote Originally Posted by matheagle View Post
    Clearly there is a typo.
    But once that is resolved, my favourite way to solve these type of limits is by multiplying by the conjugate. Not many people teach that these days.
    I was going to do that but decided it was too much work to type it up.
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  10. #10
    MHF Contributor matheagle's Avatar
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    I agree. I'm helping someone right now.
    But substituting the math and typing tex at the same is tough for me.
    I'm afraid I'll do a stupid calculation and I cannot edit after 15 minutes.
    Leaving forever with a dumb math error.
    That would be embarassing.
    I really like the conjugate technique.
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  11. #11
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    uhm... could you just simply tell me the procedure.? or just give a simple example. So that I can figure it out.... I was stuck in how to get x out of the squareroot value.
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  12. #12
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    Quote Originally Posted by thangbe View Post
    I understand in your post #2 that why I get that answer. However my teacher didn't allows us to use that menthod thank you for you help sir.
    Note that \frac{1}{\sqrt{a + x + 2}} - \frac{1}{\sqrt{a+2}} = \frac{\sqrt{a+2} - \sqrt{a+x+2}}{\sqrt{a+x+2} \, \sqrt{a+2}}.

    Therefore:



    \frac{\frac{1}{\sqrt{a + x + 2}} - \frac{1}{\sqrt{a+2}}}{x} = \frac{\sqrt{a+2} - \sqrt{a+x+2}}{x \sqrt{a+x+2} \, \sqrt{a+2}}



    = \frac{(\sqrt{a+2} - \sqrt{a+x+2}) \cdot (\sqrt{a+2} + \sqrt{a+x+2})}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}



    = \frac{(a + 2) - (a + x + 2)}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}



    = \frac{-x}{x \sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}



    = \frac{-1}{\sqrt{a+x+2} \, \sqrt{a+2} \cdot (\sqrt{a+2} + \sqrt{a+x+2})}.


    Now take the limit x \rightarrow 0.
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  13. #13
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    thank you so much for helping
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