Is $f(x,y) = (x^3 + xy^3) / (x^2 + y^6)$ differentiable at 0?

Asked Jul 31 '16 at 12:04

Active Jul 31 '16 at 12:04

Viewed 153 times

Is $f(x,y) = (x^3 + xy^3) / (x^2 + y^6)$ (for $x \neq 0$ or $y \neq 0$),

$f(0,0) = 0$

differentiable at 0?

Answer: No.

Proof: f is not continuous at 0: If we approach 0 via $x = y^3$, the limit of $f$ equals $1/2$.

Thus $f$ is not differentiable at $0$ as well.

Is my reasoning correct?

asked Jul 31 '16 at 12:04

CHwC

3

That looks correct. – marwalix Jul 31 '16 at 12:11
3

@Kenny Lau $lim_{x\to 0}\frac{x^3+x\times x}{x^2+x^2}=lim_{x\to 0}x+\frac{1}{2}=\frac{1}{2}$. I got $\frac{1}{2}$ – daruma Jul 31 '16 at 12:13

0 Answers0