9
$\begingroup$

I am wondering about a multivariable limit, and in particular, is it ever valid to use L'hospital rule.

For example, I am working on $$ \lim_{(x,y) \to (1,1)} \frac{x^3-y}{x-y}$$

This is what I have done,

let $$f(x,y)=\frac{x^3-y}{x-y}$$

$f(x,0) \rightarrow 1$ as $(x,y) \rightarrow (1,1)$

and similiary

$f(0,y) \rightarrow 1$ as $(x,y) \rightarrow (1,1)$

Okay now here is where I have a few questions ( I haven't looked at the answer or used wolfram or anything because I want to make sure I understand it first), should I continue to try out different parts, or should I try to see if I can prove the limit is 1.

in trying different paths, say $$f(x,x^2)=\frac{x^2(1-x^3)}{(1-x)}$$ would it now be valid to use L'hospital? because the y is gone and we would have 0/0 as x $\rightarrow 1$? or is it never valid to use this rule for multi valued?

Is this the right approach I should be taking or is there something else I should be thinking of?

Thank you

$\endgroup$
7
  • $\begingroup$ That approach is fine. In this case, you don't need L'Hopital, since $\frac{1-x^3}{1-x}=1+x+x^2$ when $x\neq 1$. But your formula for $f(x,x^2)$ seems wrong. $\endgroup$ Mar 23 '15 at 20:32
  • 1
    $\begingroup$ You won't dare using L'Hospital's rule for a quotient of polynomials? That's using a sledgehammer to crash a nut. $\endgroup$
    – Bernard
    Mar 23 '15 at 20:33
  • $\begingroup$ @ThomasAndrews , I see, and Im just wondering , you can only approach upon paths that are in the domain correct? $\endgroup$
    – Quality
    Mar 23 '15 at 20:33
  • $\begingroup$ The answers have addressed a few relevant points. I'll be addressing other problems. A sublimit is a limit of a restriction of a function. Something like "$f(x,0) \rightarrow 1$ as $(x,y) \rightarrow (1,1)$" would be another way of writing $$\lim \limits_{(x,y)\to (1,1)}g(x,y)=1,$$ where $g$ is $f$ restricted to $\{(x,0)\colon x\in \mathbb R\}\cap \text{dom}(f)$. But limits can only be taken at points belonging to the closure of the domain of a function and $(1,1)$ isn't in the closure of the domain of $g$. What you wrote has no meaning. $\endgroup$
    – Git Gud
    Mar 23 '15 at 20:55
  • $\begingroup$ @GitGud Thanks, but what part has no meaning, taking the limit of f(x,x^2)? $\endgroup$
    – Quality
    Mar 23 '15 at 21:08
11
$\begingroup$

Bear in mind the L'Hospital's rule goes for single-variable limits, only. Checking a lot of different paths will not guarantee the existence of the limit. But if you find any two different paths which give you different numbers, then the limit does not exists.

That being said, once you have chosen a path, the limit becomes a single-variable on, so yes, you can use L'Hospital. For example:

in trying different paths, say $f(x,x^2)=\frac{x^2(1−x^3)}{(1−x)}$ would it now be valid to use L'hospital? because the $y$ is gone and we would have $0/0$ as $x →1$?

Here you chose a path, and now you have a single-variable limit. You can use L'Hospital.


Edit: It seems that there is a sort of L'Hospital's rule for multi-variable limits, as pointed by Git Gud in the comments. Check it out.

$\endgroup$
5
  • $\begingroup$ Of course, checking a lot of paths to find a counter-example will be useful, however. :) $\endgroup$ Mar 23 '15 at 20:33
  • $\begingroup$ in which case we would have 5, implies the limit does not exist than? $\endgroup$
    – Quality
    Mar 23 '15 at 20:46
  • $\begingroup$ I don't know, haven't done the calculations. Give me 2 seconds. $\endgroup$
    – Ivo Terek
    Mar 23 '15 at 20:49
  • $\begingroup$ I got $\lim_{x \to 1}f(x,x^2)=3$, but ok, the limit won't exist anyway. $\endgroup$
    – Ivo Terek
    Mar 23 '15 at 20:50
  • $\begingroup$ Yes I meant 3 , $\endgroup$
    – Quality
    Mar 23 '15 at 21:06
0
$\begingroup$

There will be some cases where L'Hopital's rule naively applied will give the right answer for a double limit. But to understand when you are in such a case would require a deeper knowledge of the function, so l'Hopital's rule is pretty useless for proving the limit exists (although it is still useful to prove tha a limit does not exist).

$\endgroup$
0
$\begingroup$

Set $x=1+t, y=1+t^{\alpha}$. Then $$f(x,y)=\frac{t^3+3t^2+3t-t^{\alpha}}{t-t^{\alpha}}\sim_0\begin{cases}\dfrac{3t}t=3&\text{if}\enspace \alpha>1,\\\dfrac{-t^{\alpha}}{t^{\alpha}}=-1&\text{if}\enspace 0<\alpha<1\end{cases}.$$ Hence the limit is $3$ if $\alpha>1$, $-1$ if $0<\alpha<1$. This proves there is no limit at $(1,1)$.

$\endgroup$
3
  • $\begingroup$ Thanks, but where would the intuition for this come from? Does it always work? $\endgroup$
    – Quality
    Mar 23 '15 at 21:07
  • $\begingroup$ @LearningMath This question might interest you. There's no recipe for finding multivariate limits, you practice enough and think about the examples enough to get a feeling of what happens. On another note, as useful as this answer is to solving the problem, I don't think it answers any of your questions, so perhaps you might want to adjust your question to encompass Bernard's answer as a possible answer. $\endgroup$
    – Git Gud
    Mar 23 '15 at 21:10
  • 1
    $\begingroup$ The most common way to find a limit (if there is one) or prove there isn't, is to use polar coordinates and try to factor the function as the polar radius times a bounded expression. If you can't do it, you may suspect there is no limit. $\endgroup$
    – Bernard
    Mar 23 '15 at 21:59
0
$\begingroup$

enter link description hereL.Hopital rule is used in the case of indeterminate forms. the present example is suitable for existence limits at $(1,1)$ but not equal. This way, limit does not exist is the conclusion. Therefore, this example is not suitable for L.Hopital rule for multivariate function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.