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Best way of evaluating limits of multi variable functions

  1. Mar 17, 2006 #1
    I find that the best way of evaluating (non obvious) limits of multivariable functions is by using polar coords. First (if necessary) convert or reassign the function so that at the limit point, the variables all tend to 0 and than

    sub in x=rcos(angle), y=rsin(angle) where angle is arbitary.

    Than let x,y->0 as r->0. This way the limit if it exists will be found everytime because the arbitary angle will mean that graphs could come in at all possible way.

    Other ways of evluating these limits involving rearranging the function and than apply the Sandwich theorm or other ones to find the limit. Or sub different functions as they near the limit point such as linear, quadratic, hyperbola etc. These ways all involve a number of steps and involve so much extra work and guess work. Polar coordinates seem to solve the problem in one go. My Uni does not teach the polar coords way but the latter ways which is strange to me.
    Last edited: Mar 17, 2006
  2. jcsd
  3. Mar 18, 2006 #2


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    The method used depends first on whether or not you think the limit exists: if it does not exist (dne), then the method of subing-in different functions to show that the value of the limit varies depending on how the point is approached, example:

    If [itex]f(x,y)=\frac{xy}{x^2+y^2},[/itex] then [itex]\lim_{(x,y)\rightarrow (0,0)} f(x,y)[/itex] is dne since

    [tex]\lim_{x\rightarrow 0} f(x,\pm x) = \lim_{x\rightarrow 0} \frac{x(\pm x)}{x^2+(\pm x)^2}= \lim_{x\rightarrow 0} \frac{\pm x^2}{2x^2} =\pm \frac{1}{2}, [/tex]

    which implies that the value of the limit depends on the method of approach [end example.]

    Also, if after transforming to polar/sphereical coordinates (when the limit is at the origin) the function contains [tex]\theta[/tex] and/or [tex]\phi[/tex], then it (the limit) is dne.

    Otherwise, that is when the limit does seem to exist, clever use of inequalities in conjunction with the [tex]\epsilon ,\delta[/tex] definition of a limit may be employed to produce a proof; see this thread (read post #7) for an example of this technique.
  4. Mar 18, 2006 #3

    The thread you referred to is where I first learned this method of evaluating these limits. Delta and Epsilon proofs could be used but why bother when there is one system that does everything for you. Polar cords will tell you if a limit exists and what it is. It also accutrately tell you if a limit doesn't exist when you are left with angles only. It looks like the best.

    I wonder if it has any limitations.
  5. Mar 18, 2006 #4


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    Uhmm, not quite correct, if the angle, i.e [tex]\theta[/tex] appears in the numerator, then the limit can still exists. But if it appear in the denominator, then you should beware as the limit may not (I just say may not) exist there.
    Example 1:
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2y}{x ^ 2 + y ^ 2}[/tex]. After changing to polar-coordinate, we have:
    [tex]\lim_{r \rightarrow 0} r \cos ^ 2 \theta \sin \theta[/tex]. Now since:
    [tex]-1 \leq \sin \theta , \cos \theta \leq 1[/tex], so by using the Squeeze Theorem, one can show that the limit is 0, i.e:
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x ^ 2y}{x ^ 2 + y ^ 2} = 0[/tex].
    The function contains [tex]\theta[/tex] but the limit does exist.
    Example 2:
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy ^ 3}{2x ^ 2 + 3y ^ 6}[/tex].
    By changing to polar-coordinate, we have:
    [tex]\lim_{r \rightarrow 0} \frac{r ^ 2 (\cos \theta \sin ^ 3 \theta)}{2 \cos ^ 2 \theta + 3 r ^ 4 \sin ^ 6 \theta}[/tex]
    [tex]\theta[/tex] appears in the denominator, you cannot conclude anything about this limit. And you start to suspect that the limit does not exist, since if [tex]\theta \rightarrow \frac{\pi}{2}[/tex], as [tex]r \rightarrow 0[/tex], then the limit is in one of the Indeterminate forms 0 / 0.
    And so, you should look for an example that shows the limit avaluates to different values as we approach (0, 0) along different paths.
    x = 0, and x = y3 works in this case.
    Example 3:
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy}{|x| + |y|}[/tex]. Now change to polar-coordinate, we have:
    [tex]\lim_{r \rightarrow 0} \frac{r \sin \theta \cos \theta}{|\cos \theta| + |\sin \theta|}[/tex].
    One can show that:
    [tex]1 \leq | \cos \theta | + | \sin \theta | \leq \sqrt{2}[/tex], so by using the Squeeze Theorem, one can again, say that the limit is 0.
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{xy}{|x| + |y|} = 0[/tex].
    So it's not that if the expression contains [tex]\theta[/tex], or if [tex]\theta[/tex] is in the denominator then the limit does not exist, it's just one should make careful judgement to see if the limit exists or not exists. And you can improve your skills by practising...
    By the way, as you said before, the limitation of polar coordinate is that it not taught in your university...
    P.S, and also sometimes, we need to use some well-known limits of one variable function.
    Example 4:
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}[/tex]
    In this case, polar-coordinate fails. Since if we change everything to polar-coordinate, we have:
    [tex]\lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{\sin (r \cos \theta) + \sin (r \sin \theta)}[/tex], and this looks way too complicated, right? So we must find another way to go about this problem:
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}[/tex]
    [tex]= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{2 \frac{x + y}{2}}{2 \sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}[/tex]
    [tex]= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}[/tex]
    We then can use:
    [tex]\lim_{r \rightarrow 0} \frac{\sin r}{r} = 1[/tex], and conclude that:
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y} = 1[/tex]
    Last edited: Mar 18, 2006
  6. Mar 18, 2006 #5


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    Quite true, but this is easily modified:

    If the value of the limit depends on [tex]\theta[/tex] and/or [tex]\phi[/tex], the the limit is dne.

    Note that, by the definition of a limit, we have

    [tex]\lim_{(x,y)\rightarrow (a,b)}f(x,y)=L \Leftrightarrow \lim_{(u,v)\rightarrow (0,0)}f(u+a,v+b)=L\Leftrightarrow\lim_{r\rightarrow 0^+}f(a+r\cos\theta,b+r\sin\theta)=L[/tex]​

    according to my quick'n'dirty scratch work; but if, rather, I'm full of it: let me know.:biggrin:
  7. Mar 19, 2006 #6
    With example 2, since (by using polar coords) you showed that different limits are possible it is clearly indeterminate. Other methods are not necessary.

    How does your example 4 work with the substitution inside the sin and cos?
    I agree that with this one, another method must be used since polar coords will give 0/0.
  8. Mar 19, 2006 #7


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    Hmm, yeah, you may say that, but giving example is also a good way.

    SInce x, and y both tend to 0, then x - y must also tends to 0, right?
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \cos \left( \frac{x - y}{2} \right) = 1[/tex], no?
    Now split it into a product 2 of separate limits:
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}[/tex]
    [tex]= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}[/tex]
    [tex]= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right)} \times \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{1}{\cos \left( \frac{x - y}{2} \right)}[/tex]
    Now since (x, y) -> (0, 0), so x + y must also tend to 0, right?
    Using the well-known limit:
    [tex]\lim_{r \rightarrow 0} \frac{\sin r}{r} = 1[/tex], we have:
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y} = 1 \times \frac{1}{1} = 1[/tex].
    Can you get it? :)
  9. Mar 19, 2006 #8
    How did you go from:

    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}[/tex]


    [tex]= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}[/tex]
  10. Mar 19, 2006 #9
    How did you go from here:

    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}[/tex]

    to here:

    [tex]= \lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\frac{x + y}{2}}{\sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}[/tex]
  11. Mar 20, 2006 #10


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    pivoxa15, it's one of the Sum-to-product identities, you can look it un in your textbook or view it here.
    [tex]\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right)[/tex].
    Now apply that identity, and use the fact that:
    [tex]x + y = 2 \frac{x + y}{2}[/tex] to arrive at that expression.
    Can you get it? :)
  12. Mar 20, 2006 #11

    I understand. By using trig identities, you are able to evaluate the limit even when polar coords does not work.

    So there are situations where polar coords fail but other methods work. Which means that if the polar coords give an indeterminate form, it does not necessary mean the limit does not exist.
  13. Mar 21, 2006 #12


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    Uhmm,... it does work, but it looks way complicated than the way I show you above. The method is the same, by using the Sum to Product identities, we have:
    [tex]\lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{\sin (r \cos \theta) + \sin (r \sin \theta)}[/tex]
    [tex]= \lim_{r \rightarrow 0} \frac{r (\sin \theta + \cos \theta)}{2 \sin \left( r \frac{\cos \theta + \sin \theta}{2} \right) \cos \left(r \frac{\cos \theta - \sin \theta}{2} \right)}[/tex]
    [tex]= \lim_{r \rightarrow 0} \frac{r \frac{\sin \theta + \cos \theta}{2}}{\sin \left( r \frac{\cos \theta + \sin \theta}{2} \right) \cos \left(r \frac{\cos \theta - \sin \theta}{2} \right)} = 1[/tex].
    It looks much harder, right?
    By the way, you should be flexible, i.e, don't just apply polar-coordinate to solve 2 variable functions (though, it can be used on most 2-variable limit), just try every way you can think off to see if you can work out the problem.
    And again, when using polar-coordinate, the limit does not exist when the limit is dependent on [tex]\theta[/tex].
    In Example 1, 3, 4, the expression is independent of [tex]\theta[/tex], i.e [tex]\theta[/tex] can take whatever value, and it does not affect the limit.
    In Example 2, the expression is dependent on [tex]\theta[/tex].
    Can you get this? :)
    Last edited: Mar 21, 2006
  14. Mar 21, 2006 #13

    Have you left out a step? It still looks like 0/0 with the numerator going to 0 (as r->0) and denominator going to 0 (0*1=0).
  15. Mar 22, 2006 #14


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    Yes, of course, but have you looked at that closely? That can be done exactly the same as the way I've previouisly shown you.
    I left out a step to make you do some work, and think about the problem.
    You should apply the limit:
    [tex]\lim_{t \rightarrow 0} \frac{\sin t}{t} = 1[/tex].
    Apply that limit to this part:
    [tex]\frac{r \frac{\sin \theta + \cos \theta}{2}}{\sin \left( r \frac{\cos \theta + \sin \theta}{2} \right)}[/tex].
    Now, can you get it? :)
  16. Mar 22, 2006 #15
    Maybe I got it wrong but it looks to me that the limit
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x + y}{\sin x + \sin y}[/tex]
    clearly does not exist, is that your result?
    When approaching the (0,0) point from y=-x direction, the function is 0/0 and thus undefined (not indeterminate, as some of you say) - which means that there exists no neighbourhood of (0,0) such that
    1) the function is defined on the neighbourhood.
    2) the function values are "close" (closer than the gived epsilon from the limit definition) to the limit value, if any.

    which is the definition of the limit. Conclusion: the limit does not exist.
    Where am I wrong?
    Last edited: Mar 22, 2006
  17. Mar 22, 2006 #16
    VietDao, I have just realised that the function is a multivariable one either F(x,y) or F(r,theta). The limit as r->0, sin(r)/(r) -> 1 works for one variable functions because you differentiate the top and bottom to get cos(r)/1 -> 1 as r->0. But the derivative of multivariable functions are matrices and you cannot divide matrices by matrices.

    So Hooker might be right in that this functions is indeterminate after all.
  18. Mar 23, 2006 #17
    I believe that the limit indeed does not exist, one must be careful with multivariable limits, even with the most obvious limits. For example
    [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1[/tex]
    [tex]\lim_{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x}{x}[/tex]
    does not exist, for the same reasons I described in my previous post (this time when approaching (0,0) from x=0 direction - again, no neighbourhood exists...).

    But, pivoxa15, when some expression is 0/0 or whatever/0 or tan(pi/2), it's value is undefined and certainly not indeterminate. There is no nonsense like indeterminate, although it is in textbooks for centuries. I am not trying to be smarter that all of you, but it's time to realize that something either is defined or it is not and there's nothing in between. Maybe I act stupid but there are some things in math against which I fight whenever I encounter them and this is one of them, so forgive me if I am overreacting.
  19. Mar 23, 2006 #18


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    Yeah, sorry for such stupid confusion... :blush:
    hooker27, I don't understand:
    Yes, it's true that you are overreacting. whatever / 0 is not one of the Indeterminate Forms.
    Let's consider:
    0 / 0 = a. It's means that 0 = 0a. Which is a true statement, right? So when evaluating a limit, and you encounter 0 / 0 form, you don't instantly know what the answer is. It can be 1, 2, 0, or even infinity. Hence, it's the Indeterminate Forms.
    It's not like some limits as: 1 / 2, or 3 / 4, or..., all of which you can just "plug" the value of x in to get the answer. Indeterminate Forms requires you to do some manipulations before you can plug x in.
  20. Mar 23, 2006 #19
    Ok, I'll calm it down but tell me waht you think about this:

    You are right that 0=0a is true for all a but 0/0=a is nonsense since 0/0 is not assigned any value, you can't devide by 0. But I don't want to argue over something so simple, let's look at this:

    If you say, that 1/0 does not have any real value or for short you say that 1/0 is undefined, you just developed a name for some category of expressions (undefined = does not have any value). And then you trip over 0/0 and say "This I will call indeterminate!", I get the feeling that you imply that 0/0 is in some way different from 1/0 (in meaning, not subtraction), otherwise you would call it undefined as well. The only possible difference might be that you believe that under some special circumstances, 0/0 might have some real value. But this is not true!. Since the creation of the world 0/0 never had any value and it will never have any in the future, under whatever circumstances and regardless of how advanced mathematical weapon you bring to it (limits, for example). This is the reason why I don't like calling expressions like 0/0, 1^infinity etc. indeterminate because it means that you admit any difference between 1/0 and 0/0 etc. where there is none.

    Now I understand that the reason why you tend to call such expressions indeterminate is the fact that, for example
    [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1[/tex] and [tex]\lim_{x \rightarrow 0} \frac{2x}{x} = 2[/tex]
    and both expressions inside the limits are "0/0" so one gets the impression that the value of 0/0 chages depending on the context. But in the background of this "impression" is a big misunderstanding of limits. A limit [tex]\lim_{x \rightarrow a} f(x) [/tex] never looks and the funtion at the point a, whether or not the function has a value in a has absolutely no influence whatsover on the limit value so in the example above, the point 0/0 or 2*0/0 was never considered (evaluated), so you can't say that once it's value was 1 and then 2. And - logically going on - you can't give it a name ("indeterminate") depending on the result of the limit.

    You should now understand why I am against giving weird names to expressions like 0/0, tell me what your opinion is. Or forget it forever, I don't care, it's just that I'd like to see if this is making any sense to anyone except me.

    Speaking of limits, my english is pretty limited so forgive me if it's too painful to read.
    Last edited: Mar 23, 2006
  21. Mar 24, 2006 #20

    Wordings aside, I think you have to be careful with things like

    [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1[/tex] and [tex]\lim_{x \rightarrow 0} \frac{2x}{x} = 2[/tex]

    If you were in some maths or physics problem and you come to x/x would you cancel it and write 1 and in addition put a note beside it saying x cannot equal 0?

    Or would you do the cancellation to get 1 and leave it at that?

    I haven't seen anyone do the former.
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