Question

What is the limit of `lim_((x,y)->(0,0))x^3/(x^2+y^2)`?

Answer 1

`color(blue)[lim_{x->0," "y=0} x^2/(x^2+y^2)!=lim_{y->0," "x=0} x^2/(x^2+y^2)]`
Thus, the original limit does not exist.

Explanation:

In order for this limit to exist, the fraction `x^2/(x^2+y^2)`must approach the same value `L`

regardless of the path along which we approach`(0,0)`
Consider approaching`(0,0)` along the x-axis That means fixing
`y=0` and finding the limit `color(red)[lim_(x->0) x^2/(x^2+y^2)]`

`lim_{x->0," "y=0} x^2/(x^2+y^2)=lim_(x->0)x^2/(x^2+0)=0/0`

since the direct compensation product equal `0/0` we will use
L'Hôpital's rule

`lim_[(x)rarr(a)][f'(x)]/[g'(x)]`

`lim_(x->0)x^2/(x^2+0)=lim_(x->0)(2x)/(2x)=0/0`

since the direct compensation product also equal `0/0` we will use
L'Hôpital's rule again

`lim_(x->0)(2x)/(2x)=lim_(x->0)2/2=1`

Now, consider approaching,(0,0) along the y-axis This means fixing x=0 and finding the limit `color(red)[lim_(y->0) x^2/(x^2+y^2)]`

`lim_{y->0," "x=0} x^2/(x^2+y^2)=lim_(y->0)0/(0+y^2)=0/0`

`lim_(y->0)0/(0+y^2)=lim_(y->0)0/(2y)=lim_(y->0)0/2=0`

Approaching the origin along these two different paths leads to different limits

`color(blue)[lim_{x->0," "y=0} x^2/(x^2+y^2)!=lim_{y->0," "x=0} x^2/(x^2+y^2)]`

Thus, the original limit does not exist.