I'm working on partial derivatives and I'm taking calculus IV. I need some help understanding why part $ c$'s limit "does not exist".

Find each of the following limits, or explain that the limit does not exist.

Let $f(x,y) =\begin{cases} 1&, y \geq x^4\\ 1&, y \leq 0 \\ 0&, \text{otherwise} \end{cases}$

a) limit of $f(x,y)$ as $(x,y)$ approaches $(0,1)$;

b) limit of $f(x,y)$ as $(x,y)$ approaches, $(2,3)$;

c) limit of $f(x,y)$ as $(x,y)$ approaches $(0,0)$.

Is part $c$ "does not exist" because the limits of part a and b are different? Or am I wrong? Thanks.

  • $\begingroup$ no, parts a) and b) don't have anything to do with c). to prove the limit doesn't exists try to find two sequences $(x_n, y_n)$ and $(x'_n, y'_n)$ converging to $(0,0)$ along which the limits of $f$ will be different. (hint - in your case you should aim for limits $1$ and $0$ naturally) $\endgroup$
    – mm-aops
    Sep 27, 2014 at 20:15

2 Answers 2


If you approach $(0,0)$ along the path $x = 0$, then we get a limit of $1$, since we always fall under either the first or the second cases of the piecewise function.

If you approach $(0,0)$ along the path $y = 0.5x^4$, then we get a limit of $0$, since we always fall under the third case of the piecewise function.

Since different paths yielded different limits, it follows that: $$ \lim_{(x,y) \to (0,0)} f(x,y) $$ does not exist.

  • $\begingroup$ Did you just choose to use the equation y=0.5x^4? Is it possible to pick something else like y=x? $\endgroup$
    – Katie
    Sep 27, 2014 at 19:49
  • $\begingroup$ Draw a picture. You want to find a function $y = g(x)$ such that: $$ 0 \leq g(x) \leq x^4 $$ for all $(x,y)$ in some neighbourhood of $(0,0)$. Choosing $g(x) = x$ doesn't quite work. $\endgroup$
    – Adriano
    Sep 27, 2014 at 19:51
  • $\begingroup$ Oh! I see! Thanks. $\endgroup$
    – Katie
    Sep 27, 2014 at 19:52

For a limit to exist, it must be independent of the path of approach. Note that this is a necessary but not a sufficient condition for existence.

Approaching the origin along $y=x^4$ and along the path $y = ky^4 $ where $0<k<1$ yield different limits (try it), and so this particular limit cannot exist.

  • $\begingroup$ @ Karen - Because the limit is path-dependent. Limits must be path independent to exist. $\endgroup$
    – Eweler
    Sep 27, 2014 at 20:06
  • $\begingroup$ Oh, ok. Thanks for the clarification! $\endgroup$
    – Katie
    Sep 27, 2014 at 20:09

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