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Thread: show that limit exist

  1. #1
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    show that limit exist

    Let $\displaystyle f(x,y) $be defined as follows:
    $\displaystyle f(x,y) = 0$ for all (x,y) unless $\displaystyle x^4<y<x^2$
    $\displaystyle f(x,y) = 1$ for all (x,y) where $\displaystyle x^4<y<x^2$
    Show that $\displaystyle f(x,y)-->0$as $\displaystyle (x,y)-->(0,0)$ on any straight line through (0,0)
    Determine if $\displaystyle lim f(x,y)$ exist as $\displaystyle (x,y)-->(0,0)$


    just need some explaination on this question
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  2. #2
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    First check that if you approach (0, 0) along x = 0 or y = 0, then x^4 < y < x^2 is never the case, so the limit is 0.

    If the line through (0, 0) is not vertical or horizontal, then it is described by an equation y = kx for some k <> 0. Therefore, f(x, y) = 1 on this line iff x^4 < kx < x^2.

    There are several cases to consider here. Assume that x > 0. Then we can divide by x without changing the inequality: x^3 < k < x. So, $\displaystyle f(x,y(x)) = 1$ iff $\displaystyle k < x < \sqrt[3]{k}$. If k >= 1, then this is never the case, so approaching (0, 0) along such line from the right has limit 0. If k < 1, then such limit is also 0 because eventually x <= k. However, as k becomes smaller and smaller, you have f(x, y(x)) = 1 for x that are arbitrarily close to 0. This means that in every neighborhood around (0, 0) there are points (x, y) such that f(x,y) = 0 and other points (x', y') such that f(x',y') = 1.

    There may be some cases left to consider.
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