# Why does the limit not exist?

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.

• 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) Sep 27, 2014 at 20:15

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.

• Did you just choose to use the equation y=0.5x^4? Is it possible to pick something else like y=x? Sep 27, 2014 at 19:49
• 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. Sep 27, 2014 at 19:51
• Oh! I see! Thanks. 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.

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