multivariable limit: limit [(x,y)->(0,0)] [(x^2 y^2 sin(x/y)) / (x^2 + y^2)]
- Thread starter Krumzeh
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- Pick a pathway.
- See where it leads.
- Repeat as necessary.
For example, Start at (1,0) and approach along the x axis. Result?
For example, Start at (0,1) and approach along the y axis. Result?
For example, Start at (1,1) and approach along the line y = x. Result?
Proving the limit does NOT exist is easier. Proving the limit DOES exist - and what it is - takes a little more effort.
HallsofIvy
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What tkhunny suggests is that if you find that the limit, as you approach along different paths, is different, then the "limit" itself does not exist. If you believe that the limit does exist, then I would suggest changing to polar coordinates! That way the "closeness" to (0, 0) depends only on the single variable r.
Here, \(\displaystyle f(x,y)= \frac{x^2y^2sin(x/y)}{x^2+ y^2}\). In polar coordinates, \(\displaystyle x= r cos(\theta)\), \(\displaystyle y= r sin(\theta)\), \(\displaystyle sin(x/y)= sin(cos(\theta)/sin(\theta))\), and \(\displaystyle x^2+ y^2= r^2\). So the function is \(\displaystyle \frac{r^4 cos^2(\theta)sin^2(\theta) sin(cos(\theta)/sin(\theta))}{r^2}= r^2 cos^2(\theta) sin^2(\theta) sin(cos(\theta)/sin(\theta))\) all those "sin"s and "cos"s stay between -1 and 1 so that will go to 0 as r goes to 0 for any angle, \(\displaystyle \theta\).
Here, \(\displaystyle f(x,y)= \frac{x^2y^2sin(x/y)}{x^2+ y^2}\). In polar coordinates, \(\displaystyle x= r cos(\theta)\), \(\displaystyle y= r sin(\theta)\), \(\displaystyle sin(x/y)= sin(cos(\theta)/sin(\theta))\), and \(\displaystyle x^2+ y^2= r^2\). So the function is \(\displaystyle \frac{r^4 cos^2(\theta)sin^2(\theta) sin(cos(\theta)/sin(\theta))}{r^2}= r^2 cos^2(\theta) sin^2(\theta) sin(cos(\theta)/sin(\theta))\) all those "sin"s and "cos"s stay between -1 and 1 so that will go to 0 as r goes to 0 for any angle, \(\displaystyle \theta\).