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    f(x) = \frac{x_1x_2}{|x_1| + |x_2|}

    I'm asked to find the limit as x tends to 0 if it exists from the definition.

    So I need to show that, given e > 0, there exists d > 0, s.t. if ||x - 0|| < d, then ||f(x) - L|| < e where L is the limit of f(x) as x tends to 0.

    ||x - 0|| = ||x|| &lt; d

    So x_1^2 + x_2^2 &lt; d^2

    And noting that x_1x_2 &lt;= 0.5(x_1^2 + x_2^2)

    And that |x_1| +|x_2| = \sqrt{(|x_1| + |x_2|)^2} = \sqrt{|x_1|^2 + |x_2|^2 + 2|x_1x_2|} \geq \sqrt{x_1^2 + x_2^2}

    We get that

    ||f(x) - L|| = |\frac{x_1x_2}{|x_1| + |x_2|} - L| \leq |\frac{0.5(x_1^2 + x_2^2)}{\sqrt{x_1^2 + x_2^2}} - L| = | 0.5\sqrt{x_1^2 + x_2^2} - L| \leq |d/2 - L|

    Now if we choose d such that d/2 - L > 0, then |d/2 - L| = d/2 - L.

    And then if we let d = 2(L + e), we get that ||f(x) - L|| < e as required

    So surely we could have L = 1, 5, 10000 or whatever?
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    With these types of questions, it's a good idea to guess the limit first and then try to prove that the guess is correct. Now, if the limit exists, it must be independent of the path. Hence if we approach (0,0) along the x_1-axis, we see that x_2=0 and the limit will be 0. Now substitute this value for L into your argument and see if the required inequality holds.

    By the way, the function you have is not vector-valued; it's real-valued.
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    (Original post by mathprof)
    With these types of questions, it's a good idea to guess the limit first and then try to prove that the guess is correct. Now, if the limit exists, it must be independent of the path. Hence if we approach (0,0) along the x_1-axis, we see that x_2=0 and the limit will be 0. Now substitute this value for L into your argument and see if the required inequality holds.

    By the way, the function you have is not vector-valued; it's real-valued.
    Thanks for that tip about guessing limits.

    I understand that L = 0 works and is the correct limit, but why can't I let L = 1? The inequalities still hold, don't they?
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    You need the inequality to hold for all x with \lVert x \rVert &lt; \delta, which doesn't happen if L \ne 0.
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    (Original post by nuodai)
    You need the inequality to hold for all x with \lVert x \rVert &lt; \delta, which doesn't happen if L \ne 0.
    When does it stop happening though? In which line/bit of my answer? I just can't spot it anywhere.
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    (Original post by Swayum)
    When does it stop happening though? In which line/bit of my answer? I just can't spot it anywhere.
    Actually thinking about it, it's not that. It's the fact that you've said \left| \dfrac{x_1 x_2}{|x_1| + |x_2|} - L \right| \le \left| \dfrac{\frac{1}{2}(x_1^2 + x_2^2)}{\sqrt{x_1^2+x_2^2}} - L \right|. It'd be true if the mod signs were only round the first term or if L=0, but as it stands it's not necessarily true.

    To illustrate that |a| \le |b| \not \Rightarrow |a+c| \le |b+c|, consider a=0, b=-1, c=1.
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    (Original post by Swayum)
    Thanks for that tip about guessing limits.

    I understand that L = 0 works and is the correct limit, but why can't I let L = 1? The inequalities still hold, don't they?
    You're welcome.

    Limits are unique and so the epsilon-delta proof will only work, if at all, for one choice for the limit.
 
 
 
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