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Math Help - Continuous functions

  1. #1
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    Continuous functions


    JoinedAug 2012FromMariborPosts19Re: Limit Problem

    Dear All!
    Please, could anyone help me with this:
    Find number a is element of R so that the function f(x)=a; x<=0
    ( x^1/2-1)/(x^1/3-1); x>0

    is continuous in R?

    R: a=3/2

    p.s. how to get latex here? I tried to write by tags, but it just showed the source, what I wrote, and not the nice latex text!
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  2. #2
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    Re: Continuous functions

    Find the limit a=\lim_{x\to0}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}.

    Click "Reply With Quote" to see the LaTeX code and tags.
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  3. #3
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    Re: Continuous functions

    I am confused by the problem. Whatever, a is, this function will NOT be continuous at x= 1 because f(1) is not defined.
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  4. #4
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    Re: Continuous functions

    Judging by the supposed answer a = 3/2, maybe f should be given by

    f(x)=\begin{cases}a&x\le1\\(\sqrt{x}-1)/(\sqrt[3]{x}-1)&x>1\end{cases}

    and not

    f(x)=\begin{cases}a&x\le0\\(\sqrt{x}-1)/(\sqrt[3]{x}-1)&x>0\end{cases}

    because \lim_{x\to1}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}=\frac{3}{2}.
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  5. #5
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    Re: Continuous functions

    Hello!
    Please, why is lim (x to 1) from the same function 3/2?

    If we have 0^(1/2)=0
    0^(1/3)=0 so we should have: -1/-1=1, ok,

    but if we have lim(x to 1) then we should have:
    1^(1/3) =1
    1^(1/2)=1, isnt't it?
    so how do we get 3/2 from this?
    p.s. stil unclear how I can write in latex my questions, not only to quote what others wrote-also this unclear, do I have problem s with my pc?

    Happy New Year and many thanks!
    Last edited by marijakopljar; December 31st 2012 at 09:09 PM.
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  6. #6
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    Re: Continuous functions

    Quote Originally Posted by marijakopljar View Post
    Please, why is lim (x to 1) from the same function 3/2?
    If we have 0^(1/2)=0

    p.s. stil unclear how I can write in latex my questions, not only to quote what others wrote-also this unclear, do I have problem s with my pc?

    Do you understand this equality:
     \frac{{\sqrt x  - 1}}{{\sqrt[3]{x} - 1}} = \frac{{\sqrt[3]{{x^2 }} + \sqrt[3]{x} + 1}}{{\sqrt x  + 1}}~?

    As for LaTex here is the code for the above.

    [tex] \frac{{\sqrt x - 1}}{{\sqrt[3]{x} - 1}} = \frac{{\sqrt[3]{{x^2 }} + \sqrt[3]{x} + 1}}{{\sqrt x + 1}}~?[/tex] Click on the “go advanced” tab. On the toolbar you will see \boxed{\Sigma} clicking on that give the LaTeX wraps, [tex] [/tex]. The code goes between them.

    It definitely not your PC. You just have to learn the coding.

    Also if you would not use the fonts and color, it makes your posts easier to read.
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  7. #7
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    Re: Continuous functions

    Thank You, Plato!

    I dont understand this equation.
    How do we get it?

    Last edited by marijakopljar; January 1st 2013 at 05:48 AM.
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  8. #8
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    Re: Continuous functions

    \frac{\sqrt{x}-1}{\sqrt[3]{x}-1}
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  9. #9
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    Re: Continuous functions

    Quote Originally Posted by marijakopljar View Post
    Thank You, Plato!

    I dont understand this equation.
    How do we get it?


    \left(\frac{{\sqrt x  - 1}}{{\sqrt[3]{x} - 1}}\right)\frac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt[3]{{x^2 }} + \sqrt[3]{x} + 1} \right)}}{{\left( {\sqrt[3]{{x^2 }} + \sqrt[3]{x} + 1} \right)\left( {\sqrt x  + 1} \right)}}
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  10. #10
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    Re: Continuous functions

    ...
    Last edited by marijakopljar; January 1st 2013 at 08:31 AM.
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  11. #11
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    Re: Continuous functions

    I must appologise...completely new here..still unclear...except that we multiply with 1...how do we get the expression:

    \sqrt[3]{x^2}+\sqrt[3]{x}+1???
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  12. #12
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    Re: Continuous functions

    Quote Originally Posted by marijakopljar View Post
    I must appologise...completely new here..still unclear...except that we multiply with 1...how do we get the expression:

    \sqrt[3]{x^2}+\sqrt[3]{x}+1???
    Do you understand that (a^3-b^3)=(a-b)(a^2+ab+b^2)~?

    If not, then you need to review basic algebra.

    Other wise, if you do then apply it to this problem:
    (x-1)=(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)
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  13. #13
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    Re: Continuous functions

    Hello, Plato!
    Is (x^{\frac{1}{3}})^2
    not (x^{\frac{1}{3}})(x^{\frac{1}{3}})=x^{\frac{1}{9}}???
    Thank You!
    Last edited by marijakopljar; January 10th 2013 at 07:09 AM.
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  14. #14
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    Re: Continuous functions

    Quote Originally Posted by marijakopljar View Post
    Hello, Plato!
    Is (x^{\frac{1}{3}})^2
    not (x^{\frac{1}{3}})(x^{\frac{1}{3}})=x^{\frac{1}{9}}???
    Thank You!

    NO indeed.

    (x^{\frac{1}{3}})(x^{\frac{1}{3}})=x^{\frac{2}{3}}

    We add exponents.
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  15. #15
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    Re: Continuous functions

    Many thanks, clear now!
    What if the result 3/2 is wrong?
    What would be a then?
    p.s. how to find:

    \lim_{x\to\0}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}


    Would it not be 1???Thank You, Plato!
    Last edited by marijakopljar; January 10th 2013 at 07:28 AM.
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