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Math Help - please help me

  1. #1
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    please help me

    please help me to solve that
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by m777 View Post
    please help me to solve that
    Q1. find:

    <br />
\lim_{x \to \infty} [\sqrt{x^2+a x} -x]<br />

    rewrite as:

    <br />
\lim_{x \to \infty} [x \sqrt{1+a/x} -x]<br />

    or:

    <br />
\lim_{x \to \infty} x[ \sqrt{1+a/x} -1]<br />

    Which may be rewritten:

    <br />
\lim_{x \to \infty} \frac{ \sqrt{1+a/x} -1}{1/x}<br />

    and now L'Hopitals rule may be applied to give:

    <br />
\lim_{x \to \infty} \frac{ \sqrt{1+a/x} -1}{1/x}=<br />
\lim_{x \to \infty} \frac{ (1/2)(1+a/x)^{-1/2}(-a/x^2)}{-1/x^2}= \lim_{x \to \infty} (1+a/x)^{-1/2}(a/2)=a/2<br />

    RonL
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  3. #3
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    Well for Qn 2 its quite stright forward

    Just change all the x into (1)

    then you will get

    = 2.(1)

    = 2. (1) <--- (1) = 1 hence it is = 2
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  4. #4
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    [QUOTE=ThePerfectHacker;25185]Question 1, Use the standard rationalization trick,

    The function,
    f(x)=\sqrt{x^2+ax}-x
    And,
    g(x)=\frac{\sqrt{x^2+ax}-x}{1}\cdot \frac{\sqrt{x^2+ax}+x}{\sqrt{x^2+ax}+x}
    Agree for sufficiently large x, thus the limits are the same.

    Working with the second function we find,
    \frac{x^2+ax-x^2}{\sqrt{x^2+ax}+x}
    Thus,
    \frac{ax}{\sqrt{x^2+ax}-x}
    For suffiently large, i.e., x>0 we factor,
    \frac{ax}{x\sqrt{1+a/x}+x}
    Factor again,
    \frac{ax}{x(\sqrt{1+a/x}+1)}
    Thus,
    \frac{a}{\sqrt{1+a/x}+1}
    Now,
    \lim_{x\to \infty}=\sqrt{1+a/x}=1 limit compositions rule,
    Thus, by limit composition rule again,
    \frac{a}{\sqrt{1+a/x}+1}\to \frac{a}{1+1}=a/2
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  5. #5
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    Hello, m777!

    A slightly different approach to #1 . . .


    1)\;\lim_{x\to\infty}\left(\sqrt{x^2+ax} - x\right)

    Multiply top and bottom by (\sqrt{x^2+ax} + x)

    . . \frac{\sqrt{x^2+ax} - x}{1}\cdot\frac{\sqrt{x^2+ax} + x}{\sqrt{x^2+ax} + x} \;=\;\frac{(x^2 + ax) - x^2}{\sqrt{x^2+ax} + x} \;=\;\frac{ax}{\sqrt{x^2+ax} + x}


    Divide top and bottom by x:\;\;\frac{\frac{ax}{x}}{\frac{\sqrt{x^2+ax}}{x} + \frac{x}{x}} \;=\;\frac{a}{\sqrt{\frac{x^2}{x^2}+\frac{a^2}{x^2  }} + 1} \;=\;\frac{a}{\sqrt{1 + \frac{a^2}{x^2}} + 1}

    Take the limit: . \lim_{x\to\infty}\frac{a}{\sqrt{1 + \frac{a^2}{x^2}} + 1} \;=\;\frac{a}{\sqrt{1 + 0} + 1} \:=\:\frac{a}{2}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    3)\;f(x)\:=\:\left\{\begin{array}{ccc}\frac{x^2-a^2}{x-a} & 0 \leq x < a \\ a & x = a \\ 2a & x > a\end{array} \right\}

    Did you make a sketch?
    Code:
                |
                |     (a,2a)
                |       o * * * * * *
                |     *
                |   *
                | *     *(a,a)
                *
                |
        --------+-------+--------------- -
                |       a

    Rather obvious that it's discontinuous at x = a, isn't it?

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  6. #6
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    Quote Originally Posted by Soroban View Post
    Hello, m777!

    A slightly different approach to #1 . . .


    Multiply top and bottom by (\sqrt{x^2+ax} + x)

    . . \frac{\sqrt{x^2+ax} - x}{1}\cdot\frac{\sqrt{x^2+ax} + x}{\sqrt{x^2+ax} + x} \;=\;\frac{(x^2 + ax) - x^2}{\sqrt{x^2+ax} + x} \;=\;\frac{ax}{\sqrt{x^2+ax} + x}


    Divide top and bottom by x:\;\;\frac{\frac{ax}{x}}{\frac{\sqrt{x^2+ax}}{x} + \frac{x}{x}} \;=\;\frac{a}{\sqrt{\frac{x^2}{x^2}+\frac{a^2}{x^2  }} + 1} \;=\;\frac{a}{\sqrt{1 + \frac{a^2}{x^2}} + 1}

    Take the limit: . \lim_{x\to\infty}\frac{a}{\sqrt{1 + \frac{a^2}{x^2}} + 1} \;=\;\frac{a}{\sqrt{1 + 0} + 1} \:=\:\frac{a}{2}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Did you make a sketch?
    Code:
                |
                |     (a,2a)
                |       o * * * * * *
                |     *
                |   *
                | *     *(a,a)
                *
                |
        --------+-------+--------------- -
                |       a
    Rather obvious that it's discontinuous at x = a, isn't it?
    Unless a=0.

    RonL
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  7. #7
    Forum Admin topsquark's Avatar
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    Is a function that is only defined for a single point continuous?

    -Dan
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  8. #8
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    Quote Originally Posted by topsquark View Post
    Is a function that is only defined for a single point continuous?
    No! by definition,
    \lim_{x\to c}f(x)=f(c)
    Now in order to take a limit we need to find a \delta>0 such that,
    |x-c|<\delta is in the domain, that means,
    c-\delta<x<c+\delta
    Is some open interval.
    That means the function needs to be definied on some open interval about x=c which it fails heir.
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  9. #9
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    please help me

    Dear,
    soroban.
    In question no three i make a sketch of it.yes it is discontinuous at x=a.
    many best wishes for you
    from
    m777
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