# If a 2-dimensional limit does not exist, can we argue that the limit is greater than some minimum?

If we have a limit, for example, $$$$\lim_{(x,y) \rightarrow (0,0)} \frac{a x^2 + b y^2}{x^2 + y^2}$$$$ then we can show that the limit "does not exist" if $$a \neq b$$ since taking the limit as $$y \rightarrow 0$$ first does not yield the same result as taking the limit as $$x \rightarrow 0$$ first.

But if we let $$c > \max[a,b]$$, can we state that $$$$\lim_{(x,y) \rightarrow (0,0)} \frac{a x^2 + b y^2}{x^2 + y^2} \leq c$$$$

If we take the limit along the direction $$x=y$$ for example, the limit approaches $$\frac{a+b}{2} < c$$.

If this does work, does it work in general?

• How can you write down the limit in an inequality if it doesn't exist? Feb 4 at 14:15
• If you think about it in a 1-dimensional sense, if I have a function $y = 1$ if $x < 0$ and $y = 2$ if x > 0, then in general y > 0, is that correct? Does this includes in the limit as x goes to 0? That's what I'm asking in a multivariable case. Feb 4 at 14:18
• All the limit on two roads $A=\{(x,0):x\in\mathbb R:\}$ y $B=\{(0,y):x\in\mathbb R:\}$, llegaras que en $\displaystyle\lim_{(x,y)\in A}f(x,y)=a$ and $\displaystyle\lim_{(x,y)\in B}f(x,y)=b$. Feb 4 at 14:22

A problem I see with $$\lim_{(x,y) \rightarrow (0,0)} \frac{a x^2 + b y^2}{x^2 + y^2} \stackrel? \leq c$$ is that for a sentence of this form to be meaningful, the thing on the left side of "$$\leq$$" has to be a number. The "$$\leq$$" symbol cannot "reach inside" the limit notation to make statements about the ways in which the limit fails.

And the limit on the left hand side really does not exist in any mathematical sense. You cannot even say the limit depends on which line you approach $$(0,0)$$ along, because the concept of $$\lim_{(x,y) \rightarrow (0,0)}$$ does not say that you approach $$(0,0)$$ along a line -- you could go in along a spiral or on a path that zigzags.

But I think you could write $$\limsup_{(x,y) \rightarrow (0,0)} \frac{a x^2 + b y^2}{x^2 + y^2} = \max\{a,b\}.$$

You can just write "$$=$$" here and not "$$\leq$$" because the "$$\limsup$$" notation already incorporates the "$$\leq$$" idea. In fact it seems to be what you were trying to write with "$$\lim$$" and "$$\leq$$".

You could also write $$\frac{a x^2 + b y^2}{x^2 + y^2} \leq \max\{a,b\} \quad \text{for} \quad (x,y)\neq(0,0).$$

You don't need any kind of limit here because the statement starts out true and remains true no matter how you move $$(x,y)$$ around $$(0,0)$$.

• I see; so it would be more of a notation issue that I'm struggling with. Thanks for that Feb 4 at 14:27
• A thing about math is, if there's a useful idea that seems like it should be expressible in notation, someone has probably invented a notation to express it. Feb 4 at 14:30
• I'd add that sometimes we do have notation that "reaches inside" and changes the meanings of other symbols. E.g. usually "$\lim_{n\to\infty}f(n)=c$" and "$\lim_{n\to\infty}f(n)=\infty$" are defined separately and not interpreted as literal equalities, and "$\lim_{n\to\infty}f(n)$ does not exist" is defined even though "$T$ does not exist" is nonsense for an arbitrary term $T$. Another example is "$f(x)=x^3+o(x)$". Students who expect notation to be strictly compositional can be confused when encountering things like this. (I feel like there's a better example that I'm forgetting.)
– Karl
Feb 4 at 15:57