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Thread: How to prove multivariable limits by the epsilon delta definition

  1. #1
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    How to prove multivariable limits by the epsilon delta definition

    How to prove for example that $\displaystyle \lim_{(x,y)\to(1,1)}(x^2+y^2)=2$ ?

    Or $\displaystyle \lim_{(x,y)\to(0,1)}\frac{x+1}{y+1}=\frac12$ ?

    I want to learn it. I've been looking for some notes that show how to prove multivariable limits by the epsilon-delta definition, but not succeed.
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  2. #2
    MHF Contributor arbolis's Avatar
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    For 1), you have to find a $\displaystyle \delta (\varepsilon)$ for any $\displaystyle \varepsilon>0 $ such that if $\displaystyle |(x,y)-(1,1)|<\delta (\varepsilon) \Rightarrow |x^2+y^2-2|<\varepsilon$.
    Hint : How do you calculate $\displaystyle (x,y)-(1,1)$?
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  3. #3
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    well that'd be $\displaystyle (x-1,y-1)$ but i don't know what's next.
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  4. #4
    MHF Contributor arbolis's Avatar
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    $\displaystyle |(x,y)-(1,1)|=|\sqrt{(x-1)^2+(y-1)^2}|=\sqrt{(x-1)^2+(y-1)^2}$.

    Now you have to show that if $\displaystyle \sqrt{(x-1)^2+(y-1)^2}<\delta(\varepsilon)$ then $\displaystyle |x^2+y^2-2|<\varepsilon$.

    Hint : Try to chose a value for $\displaystyle \delta(\varepsilon)$, note that it depends on $\displaystyle \varepsilon$. For instance I could try the proof with $\displaystyle \delta(\varepsilon)=\frac{1}{\varepsilon}$. If it doesn't work I'd try with another value.
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  5. #5
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    Hello,

    As a sidenote... : you're currently working on $\displaystyle \mathbb{R}^2$
    Since we're in finite dimension, all the norms are equivalent.
    So you can choose whatever norm you want :
    norm 1 : $\displaystyle \|(a,b)\|=|a|+|b|$
    norm 2 : $\displaystyle \|(a,b)\|=\sqrt{a^2+b^2}$

    and by noting (a,b)-(c,d)=(a-c,b-d)

    maybe the norm 1 would be easier to manipulate
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