Question

limits of two variables

Answer 1

lim (x,y)→(0,0)
xy^3/( x2+y6 )


using polar coordinates i got zero
but its dne...

Answer 2

Not gonna lie, I came here because the name was "dbwong" which reminded me of the Chinese food restaurant the next town over "Debbie Wong"...so my stomach led me here

Answer 3

But in regards to the question... when faces with limits involving 2 variables, you really just want to make sure the limit is the same no matter which PATH you take to get there

If you follow this function from the 'x-axis' where y=0....what value do you get?
If you then do it from the 'y-axis' where x=0...what value do you get?
If you THEN follow it from the line y=x...what value do you get?

Etc...and you would keep checking paths like this to make sure they all yield the same result...if ANY path gives you a different result from any other path...the limit does not exist

*Those 3 specific paths were chosen as they are the 1) Easiest and 2) The most common you'll use

Answer 4

So here...

\[\large \lim_{(x,y)\rightarrow (0,0)} \frac{xy^3}{(x^2 + y^6)}\]

When following from the x-axis...x=x and y=0 so

\[\large \lim_{(x,y)\rightarrow (0,0)} \frac{x(0)^3}{(x^2 + (0)^6)} = 0\]

Now let's try it from the y-axis, x=0 and y=y

\[\large \lim_{(x,y)\rightarrow (0,0)} \frac{(0)y^3}{((0)^2 + y^6)}=0\]

Okay...now lets try y=x

\[\large \lim_{(x,y)\rightarrow (0,0)} \frac{x(x)^3}{(x^2 + (x)^6)}=\frac{x^4}{x^2(1+x^4)}=\frac{x^2}{1+x^4}\]

Uh oh...Need to get bailed out now XD Still have 0

Answer 5

Lets try another path...lets do \(\large y=x^\frac{1}{3}\)

\[\large \lim_{(x,y)\rightarrow (0,0)} \frac{x(x^{\frac{1}{3}})^3}{(x^2 + (x^{\frac{1}{3}})^6)}\]

\[\large \lim_{(x,y)\rightarrow (0,0)} \frac{x^2}{x^2 + x^2} = \frac{1}{2}\]
THERE we go...just kept following different paths but eventually we got there...since this 1/2 is different from the 0's we got before...we know the limit does not exist