Since I began studying limits of multi-variable functions, I have been baffled with this question: how can one tells if a limit exists or not? I don't know if it's the right way to solve this kind of problem but I've always identified if the limit exists first before trying to approve/disprove its existence.

For example, from the exercises I have done, I can see that those functions having this form $\frac{{{x^a}{y^b}}}{{{x^{ma}} + n{y^{mb}}}}$ will never reach a limit as $(x,y) \to (0,0)$. This will save me from computing a non-existent limit in vain.

Sorry if this question sounds dumb to you, I just don't want to screw up my upcoming test. Any tip or suggestion would be much appreciated, thanks!


1 Answer 1


In general, it is much easier to show that a limit does not exist than it is to show a limit does exist. There are a few common ways of working with multi-variable functions to obtain the existence or nonexistence of a limit:

  1. Try different paths. That is, parameterize $x$ and $y$ as $x = x(t)$, $y = y(t)$ such that $(x(0),y(0)) = (a,b)$, where $(a,b)$ is the point you want to approach in the limit. This is usually the first resort, and if the paths are chosen judiciously, you will obtain two different answers, which implies the nonexistence of the limit, because for the limit to exist, it must have the same value along every possible path. Note that this test can only be used to show nonexistence: to prove a limit exists requires more work.
  2. Use polar coordinates. This approach can prove that the limit exists in special cases. We write $x = r\cos\theta$, $y = r\sin\theta$, and as the limits are usually taken as $(x,y)\to (0,0)$, we now must look at what happens as $r\to 0^+$. Sometimes, this will depend on $\theta$, which corresponds to a specific path ($\theta$ controls the direction), and sometimes, the $r$ will dominate and leave you with an expression where $\theta$ does not matter - in this case, the limit exists. However, one must be careful, because there are some expressions that might seem to be independent of $\theta$ as $r\to 0^+$, but are not: for example, take $$ r\frac{\cos^2\theta\sin^2\theta}{\cos^3\theta + \sin^3\theta}. $$ For any constant $\theta$ such that the denominator exists (and is nonzero), the limit as $r\to 0^+$ is $0$, but there are certain paths $\theta = \theta(r)$ along which the value of the limit will not be $0$.
  3. $\delta - \epsilon$ proofs: When correct, these show the existence of a limit. However, one must already know what the limit is before this type of proof is possible. If you are unfamiliar with $\delta - \epsilon$ arguments, the statement is: Given a function $f : \mathbb{R}^n\to\mathbb{R}$, we say $$ \lim_{\vec{x}\to\vec{x}_0} f(\vec{x}) = L $$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $\left|\,f(\vec{x}) - L\right| < \epsilon$ whenever $d(\vec{x}, \vec{x}_0) < \delta$. Here, $d(\vec{x}, \vec{x}_0)$ refers to the distance in Euclidean $n$-space (for example, with $n = 2$ we have $d((a,b), (c,d)) = \sqrt{(a - c)^2 + (b - d)^2}$.) This approach should be used if you are already convinced that the limit exists and is equal to $L$.
  4. Use algebra and theory. You have probably seen that the product and sum of continuous functions are continuous, and that for continuous functions, the limit can be evaluated by "plugging in." If you can identify that your function is continuous, or at least becomes continuous after algebraic manipulation (e.g. canceling a "bad factor" from the denominator), you can use the theorems to say that the limit exists.
  • $\begingroup$ Very useful! +1 $\endgroup$ Commented Oct 11, 2017 at 6:06

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