# show that limit exist

• Oct 1st 2010, 12:02 PM
450081592
show that limit exist
Let $f(x,y)$be defined as follows:
$f(x,y) = 0$ for all (x,y) unless $x^4
$f(x,y) = 1$ for all (x,y) where $x^4
Show that $f(x,y)-->0$as $(x,y)-->(0,0)$ on any straight line through (0,0)
Determine if $lim f(x,y)$ exist as $(x,y)-->(0,0)$

just need some explaination on this question
• Oct 1st 2010, 02:53 PM
emakarov
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, $f(x,y(x)) = 1$ iff $k < x < \sqrt[3]{k}$. 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.