Limit of x^2y^2/(x^2y^2+(x-y)^2) Does Not Exist

Find the path that switches off a term

The denominator is a sum of two nonnegative pieces, $x^2y^2$ and $(x-y)^2$. Whenever the second piece is much larger than the first, the fraction is close to zero. So look for a path that makes $(x-y)^2$ disappear.

That path is the diagonal $y = x$, where $x - y = 0$ exactly.

On the $x$-axis the numerator vanishes instead, and the denominator does not.

The two values are $1$ and $0$, so the limit does not exist. Notice how sharp the split is: on the diagonal the function sits at $1$ the whole way in, while on every other line it falls away to $0$, no matter how near the origin you look.

Every other line gives zero

Substitute $y = mx$ with $m \neq 1$ and cancel a factor of $x^2$.

As $x \to 0$ the numerator goes to $0$ and the denominator goes to the positive constant $(1-m)^2$, so the value is $0$. Only the slope $m = 1$ makes that constant vanish, and there the fraction becomes $1$ instead.

• $m = 0$ gives $0$, the $x$-axis.
• $m = -1$ gives $0$, the anti-diagonal.
• $m = 2$ gives $0$.
• $m = 1$ gives $1$, the single exceptional direction.

One direction out of infinitely many carries a different value, which is enough. A limit must be blind to the route taken.

The mistake: sampling lines at random and calling it done

A student who tries $y = 0$, $x = 0$, $y = 2x$, and $y = -x$ gets $0$ four times and confidently writes limit $= 0$. The one path that matters was skipped. When the denominator contains a squared difference like $(x-y)^2$, the curve that annihilates it is always worth trying first.

The habit worth building is to read the denominator before choosing paths. Ask two questions: where can one of its terms be switched off, and where can two terms be made the same size. The first gives $y = x$ for $(x-y)^2$ and either axis for $x^2 + y^2$; the second gives $y = x^2$ for $x^4 + y^2$. Those are the paths that break limits.

It also pays to test the curve $y = x + x^2$, which approaches the diagonal without lying on it. There $f = \frac{x^2(x+x^2)^2}{x^2(x+x^2)^2 + x^4}$, and dividing through by $x^4$ gives $\frac{(1+x)^2}{(1+x)^2 + 1} \to \tfrac{1}{2}$, yet another value.