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Multivariable limit

  1. Mar 1, 2015 #1
    How do I justify that [itex]lim_{(x,y)\to (0,0)} \cos{\frac{x}{\sqrt{y}}} = 1[/itex]?
    If I approach from the y axis, it would become [itex]lim_{y\to 0} \cos{\frac{0}{\sqrt{y}}} = 1 [/itex], but if I approach from the x axis, it would become [itex]lim_{x\to 0} \cos{\frac{x}{\sqrt{0}}} = D.N.E[/itex], no? (does not exist)

    Wolfram thinks the limit is 1 no matter what and I tend to trust it, because Wolfram seems to be quite capable, but but...I am having trouble understanding how this particular limit works.
    In a way I would understand, as the distance from origin is approaching 0, then the cosine of it would be closer and closer to 1, but how do I get rid of the part where there is division by 0 , it hurts my eye(s).
     
    Last edited: Mar 1, 2015
  2. jcsd
  3. Mar 1, 2015 #2

    wabbit

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    Well you don't justify it, because it's incorrect, for the reason your example very clearly shows.
     
  4. Mar 2, 2015 #3
    Okay I was playing around with the problem a bit and augemented it with some random rational:
    [tex]\lim_{(x,y)\to(0,0)} \frac{x^4 + y^4}{x^2+2y^2}\cos{\frac{x}{\sqrt{y}}} = 0[/tex] according to Wolfram Alpha. This again leads me to believe Wolfram Alpha considers the limit of the cosine to be something specific no matter the approach direction.
    Interestingly, the limit does not exist to WA, either if:
    [tex]\lim_{(x,y)\to(0,0)} \frac{x^2 + y^2}{x^2+2y^2}\cos{\frac{x}{\sqrt{y}}}[/tex]
    A property of limits states that [itex]\exists\lim f\cdot g \Leftrightarrow \exists \lim f \wedge \exists \lim g[/itex]. Is WA breaking the rules here or is there something present that is elusive to my tiny brain?
    EDIT: MY brain is tiny - it only works [itex]\exists\lim f\cdot g \Leftarrow \exists \lim f \wedge \exists \lim g[/itex]


    Who do I believe, WA or myself? I am inclined to believe the limit does not exist as the cosine messes things up.
     
    Last edited: Mar 2, 2015
  5. Mar 2, 2015 #4

    wabbit

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    But this is different. This limit is indeed zero, for a reason that has little to do with the cosine - can you see it ?
     
  6. Mar 2, 2015 #5
    well the numerator "outpowers" the denominator, so that would be 0, but is it correct to state 0 * undefined = 0?
     
  7. Mar 2, 2015 #6

    wabbit

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    No, that's not enough.
    Hint : show that ## x^4+y^4\leq (x^2+2y^2)^2 ##
    Edit : then combine that with ## |cos(...)|\leq 1 ## and you're done
     
    Last edited: Mar 2, 2015
  8. Mar 2, 2015 #7
    [itex]x^4 + 2x^2 y^2 + y^4 \leq x^4 + 6x^2y^2 + 4y^4 [/itex]
    where does this come from, though? If you show this inequality, what exactly can you state?
    and why did you square the denominator, why not cubed or 4th or 5h? What's the logic here?

    I can't see where you come up with the 0 as the limit.
    I can agree the [itex]|\cos{f}| \leq 1 \forall f[/itex], but where does that leave the prior? If we show the inequality and then state [itex]x^4+y^4 \geq x^2 + 2y^2[/itex]?? What if it's equal? you cancel them out and get 1 * undefined = undefined.

    Please, explain more thoroughly, I cannot follow your logic at this time.

    Or perhaps you mean to show the numerator goes to zero faster than the denominator in which case its limit is 0?
     
    Last edited: Mar 2, 2015
  9. Mar 2, 2015 #8

    wabbit

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    You should look at this more carefully, I am sure you can sort it out.
    Note that the inequality you start with in your last post is not the one I quoted.
    Hint : Follow the hint.
     
  10. Mar 2, 2015 #9
    Well the only thing I can see at the moment is that you want to show the numerator goes to zero faster than the denominator. What I don't understand why you compare the numerator to the denominator squared, why not denominator cubed?

    Anyway, meanwhile I thought of using polar coordinates instead and see where that takes me.
    [itex]\lim_{r\to 0} \frac{r^4(\cos^4{\varphi} + \sin^4{\varphi})}{r^2(\cos^2{\varphi} + 2\sin^2{\varphi})}\cdot \cos{ \frac{r\cos{\varphi}} {\sqrt{r\sin{\varphi}}} } [/itex] and we see that we are left with r2 and the second factor is a cosine of 0. Limit is 0.

    If I use polar coodinates for just the cosine part [itex]\lim_{r\to 0} \frac{r\cos{\varphi}} {\sqrt{r\sin{\varphi}}} = 1 [/itex]. Why does the non-polarcoordinate approach end in conflict?

    My question still remains, though - do I have to specify that the sine of phi cannot be 0? We are multiplying by 0 so anything times 0 ought to be 0. Is it also true for 0 * something / 0?
    EDIT: I think not since that would be stating something like [itex]0\cdot\frac{A}{0} = A\cdot\frac{0}{0} = 0[/itex]
     
    Last edited: Mar 2, 2015
  11. Mar 2, 2015 #10

    wabbit

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    OK, you're getting somewhere:)
    Using polar coordimates here isn't the simplest way but it's a reasonable approach. Let's look at this, but first
    No deep reason, just because (a) the inequality does hold, while it would not be true for the cube, (b) it's easy to establish, and (c) it proves the result you seek.

    Now,
    Correct, although you still need to explain why the sine and cosine terms in the denominator do not mess things up. After all, what if ##\cos^2{\varphi} + 2\sin^2{\varphi}## were to take the value 0? You need to show that this is always greater than some fixed value, so it doesn't become troublesome. (Hint: use ##\cos^2{\varphi} + \sin^2{\varphi}=1##).

    It doesn't. In fact it's easier: by inspection,
    ##x^4+y^4\leq x^4+4x^2y^2+4y^4=(x^2+2y^2)^2##
    So
    ## \frac{x^4+y^4}{x^2+2y^2}\leq x^2+2y^2##.
    The cosine term is less than one, and you can conclude.

    You are right to worry about that. Your conclusion holds for the limit only if it holds for all values of phi, including 0.
     
    Last edited: Mar 2, 2015
  12. Mar 2, 2015 #11

    wabbit

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    You are of course correct on this, and if WA is in fact stating the equivalence, then WA is wrong.
     
  13. Mar 2, 2015 #12
    I am not concerned about the polar denominator being 0 since we are adding 2 non-negative values, where if the sine is 0 then cosine is always different from 0 and vica versa. The sum is strictly positive for every phi hence /0 is avoided.

    New problem is that you said, the limit exists iff the limit yields the same result for every phi - I should clearly state the sine of phi cannot be 0, which means there exist phi for which the limit is undefined. But cosine is fluctuating between -1 and 1 so does it really matter that there is division by 0 inside the cosine? While I would argue that [itex]0\cdot\frac{A}{0}[/itex] is undefined, I would say that [itex]0\cdot x = 0 \forall x\in [-1,1][/itex]
     
  14. Mar 2, 2015 #13

    wabbit

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    Yes, you would be correct in arguing that. Somewhat more formally, if ## f\rightarrow 0## and ## |g|\leq C ## for some constant ## C ##, then ## fg\rightarrow 0 ##. This is indeed the argument you need to use for the cosine part, so it looks like you're done : )

    Edit : however your first statement about "the denominator is not zero so the issue is avoided" is not sufficient : if the denominator approached zero even without reaching it, you'd still have a problem.
    It may seem unnecessarily bothersome, but you will not arrive at solid conclusions if you are not careful in your arguments, especially when your intuition about the topic isn't yet well developped.
     
    Last edited: Mar 2, 2015
  15. Mar 2, 2015 #14
    Well, using the trigidentity I can state that [itex]\cos^2{\varphi} + 2\sin^2{\varphi} = 2 - \cos^2{\varphi}\in [1,2], \forall\varphi[/itex] hence it can never approach zero.
    I agree, though, it's always safer to follow the definitions to the letter and leave nothing unchecked, sometimes I just get lazy :/
     
    Last edited: Mar 2, 2015
  16. Mar 2, 2015 #15

    wabbit

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    That's the spirit : )
     
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