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Math Help - Multivariable Limit Problem

  1. #1
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    Multivariable Limit Problem

    I am having some problems with the following problems:
    1. \lim_{\theta\to 0}\frac{sin\theta}{\theta}<br />

    Using
    l'H˘pital's rule I got that the limit equals.

    2. \lim_{(x,y)\to (0,0)}\frac{sin(x + y)}{x + y}<br />

    I am not really sure how to proceed here. I mean, I could approach along the y axis, then am I able to use
    l'H˘pital's rule?Also, if I approach along y = x afterward I believe I would get 2 for the limit, whereas along the y-axis would yield 1 provided I could use l'H˘pital's rule. In that case the limit would not exist.

    3. \lim_{(x,y)\to (0,0)}\frac{sin(xy)}{xy}<br />

    On this one I don't really know what to do.
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  2. #2
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    Quote Originally Posted by Alterah View Post
    I am having some problems with the following problems:
    1. \lim_{\theta\to 0}\frac{sin\theta}{\theta}<br />

    Using
    l'H˘pital's rule I got that the limit equals.

    2. \lim_{(x,y)\to (0,0)}\frac{sin(x + y)}{x + y}<br />

    I am not really sure how to proceed here. I mean, I could approach along the y axis, then am I able to use
    l'H˘pital's rule?Also, if I approach along y = x afterward I believe I would get 2 for the limit, whereas along the y-axis would yield 1 provided I could use l'H˘pital's rule. In that case the limit would not exist.

    3. \lim_{(x,y)\to (0,0)}\frac{sin(xy)}{xy}<br />

    On this one I don't really know what to do.
    1. L'Hopital's rule allows you to take \lim_{\theta\to0}\frac{\cos\theta}{1}, which is clearly 1.

    2 & 3. The rule: \lim_{anything\to0}\frac{\sin(anything)}{anything}  =1. To cover all possible directions at once (except x=0, which you can check manually), simply let y=f(x). Since f(x) must pass through the origin, f(0)=0, so both x+f(x) and x\cdot f(x) approach 0 as x\to 0 Therefore both questions fall under this rule and have limits of 1.
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  3. #3
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    Quote Originally Posted by Alterah View Post
    I am having some problems with the following problems:
    1. \lim_{\theta\to 0}\frac{sin\theta}{\theta} <br />

    Using l'H˘pital's rule I got that the limit equals.

    2. \lim_{(x,y)\to (0,0)}\frac{sin(x + y)}{x + y} <br />

    I am not really sure how to proceed here. I mean, I could approach along the y axis, then am I able to use
    l'H˘pital's rule?Also, if I approach along y = x afterward I believe I would get 2 for the limit, whereas along the y-axis would yield 1 provided I could use l'H˘pital's rule. In that case the limit would not exist.

    3. \lim_{(x,y)\to (0,0)}\frac{sin(xy)}{xy} <br />

    On this one I don't really know what to do.
    I wouldn't use L'Hospital's Rule to solve 1.

    This is because, to find the derivative of \sin{x}, you need to know the limit \lim_{x \to 0}\frac{\sin{x}}{x}, so you have a circular argument.

    You can use the sandwich theorem instead - there is plenty of information you can find on google regarding this limit.
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  4. #4
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    Quote Originally Posted by redsoxfan325 View Post
    1. L'Hopital's rule allows you to take \lim_{\theta\to0}\frac{\cos\theta}{1}, which is clearly 1.

    2 & 3. The rule: \lim_{anything\to0}\frac{\sin(anything)}{anything}  =1. To cover all possible directions at once (except x=0, which you can check manually), simply let y=f(x). Since f(x) must pass through the origin, f(0)=0, so both x+f(x) and x\cdot f(x) approach 0 as x\to 0 Therefore both questions fall under this rule and have limits of 1.
    Thanks...I forgot to put that I got the limit equals 1 for question 1. I am trying to follow what you have for part two and three. I see how it can work for part 2. But it seems for part three if we use y = f(x) I feel like we get something along the lines of sin(x*f(x))/(x*f(x)) and we still wind up with zero in the denominator. As far as not using L'Hopital's rule for part 1, I Don't really see why not. It's an indeterminate form of 0/0. So I can apply it to it.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Alterah View Post
    Thanks...I forgot to put that I got the limit equals 1 for question 1. I am trying to follow what you have for part two and three. I see how it can work for part 2. But it seems for part three if we use y = f(x) I feel like we get something along the lines of sin(x*f(x))/(x*f(x)) and we still wind up with zero in the denominator.
    We do get \frac{\sin(x\cdot f(x))}{x\cdot f(x)}. But that's good, because x\cdot f(x)\to 0 as x\to 0, which means we can apply that rule I stated.

    Quote Originally Posted by Alterah View Post
    As far as not using L'Hopital's rule for part 1, I Don't really see why not. It's an indeterminate form of 0/0. So I can apply it to it.
    What Prove It is saying is that in order to take the derivative of \sin x at 0 using the definition of the derivative, you need to take that limit.

    \lim_{h\to 0}\frac{\sin(h)-\sin(0)}{h-0} = \lim_{h\to0}\frac{\sin h}{h}
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    Quote Originally Posted by redsoxfan325 View Post
    We do get \frac{\sin(x\cdot f(x))}{x\cdot f(x)}. But that's good, because x\cdot f(x)\to 0 as x\to 0, which means we can apply that rule I stated.
    Ok. Thanks for the help, how did the rule come about? I suppose I am wanting to verify for myself that this rule holds. At this point the rule seems like "hand waving." Thanks again.
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  7. #7
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Alterah View Post
    Ok. Thanks for the help, how did the rule come about? I suppose I am wanting to verify for myself that this rule holds. At this point the rule seems like "hand waving." Thanks again.
    L'Hopital's Rule shows that it works, though you probably wouldn't want to use it in a proof, for the reasons discussed above.

    \lim_{anything\to0}\frac{\sin(anything)}{anything}  =\lim_{anything\to0}\frac{\cos(anything)\cdot\frac  {d}{dx}[anything]}{\frac{d}{dx}[anything]} =\lim_{anything\to0}\frac{\cos(anything)}{1}, which is clearly 1.

    More officially, you'd probably use something like u(x) instead of "anything", i.e.

    \lim_{u(x)\to0}\frac{\sin(u(x))}{u(x)}=1

    but I used "anything" to try to make it easier to understand.
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  8. #8
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    Quote Originally Posted by redsoxfan325 View Post
    L'Hopital's Rule shows that it works, though you probably wouldn't want to use it in a proof, for the reasons discussed above.

    \lim_{anything\to0}\frac{\sin(anything)}{anything}  =\lim_{anything\to0}\frac{\cos(anything)\cdot\frac  {d}{dx}[anything]}{\frac{d}{dx}[anything]} =\lim_{anything\to0}\frac{\cos(anything)}{1}, which is clearly 1.

    More officially, you'd probably use something like u(x) instead of "anything", i.e.

    \lim_{u(x)\to0}\frac{\sin(u(x))}{u(x)}=1

    but I used "anything" to try to make it easier to understand.
    I see...I suppose it's a bit of a challenge to see with multivariable functions because we actually describe partial derivatives. It makes absolute sense with f(x). Alternatively using the rule I'd think you can get the following:

    \lim_{u(x)\to0}\frac{\cos(u(x))}{u(x)}=0

    correct?
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  9. #9
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    Quote Originally Posted by Alterah View Post
    I see...I suppose it's a bit of a challenge to see with multivariable functions because we actually describe partial derivatives. It makes absolute sense with f(x). Alternatively using the rule I'd think you can get the following:

    \lim_{u(x)\to0}\frac{\cos(u(x))}{u(x)}=0

    correct?
    NO!

    l'Hopital's can only be used when your limit has the indeterminant form of either \frac{0}{0} or \frac{\infty}{\infty}. I doubt \lim_{u(x)\to 0}\frac{\cos(u(x))}{u(x)} will have either of these two forms, regardless of what u(x) is ....
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    NO!

    l'Hopital's can only be used when your limit has the indeterminant form of either \frac{0}{0} or \frac{\infty}{\infty}. I doubt \lim_{u(x)\to 0}\frac{\cos(u(x))}{u(x)} will have either of these two forms, regardless of what u(x) is ....
    True...I completely forgot about that when I said that about cosine.
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  11. #11
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    Quote Originally Posted by Alterah View Post
    I am having some problems with the following problems:
    1. \lim_{\theta\to 0}\frac{sin\theta}{\theta}<br />

    Using
    l'H˘pital's rule I got that the limit equals.

    2. \lim_{(x,y)\to (0,0)}\frac{sin(x + y)}{x + y}<br />

    I am not really sure how to proceed here. I mean, I could approach along the y axis, then am I able to use
    l'H˘pital's rule?Also, if I approach along y = x afterward I believe I would get 2 for the limit, whereas along the y-axis would yield 1 provided I could use l'H˘pital's rule. In that case the limit would not exist.

    3. \lim_{(x,y)\to (0,0)}\frac{sin(xy)}{xy}<br />

    On this one I don't really know what to do.
    First I will talk about 2 and 3, which follows from 1.
    Clearly, for 2, if (x,y)\to(0,0), then x+y\to0, and similarly for 3.
    Now return to 1.
    Let C be the unit ball (hence its area is \pi).
    Draw an equilateral triangle in C such that its corners touch C, denote by A_{3} the area of the triangle.
    Then A_{3}=3\frac{1}{2}\sin\bigg(\frac{2\pi}{3}\bigg).

    Now draw a square in C, again let it corners touch C, denote by A_{4} the area of the square.
    Then A_{4}=4\frac{1}{2}\sin\bigg(\frac{2\pi}{4}\bigg).

    Similarly, draw a polygon (with n edges of the same lenght), and by A_{n} denote the area of the polygon.
    As the number of the corners (edges) increase, the area of the polygon tends to the area of the unit ball (exactly \pi), i.e.,
    \pi=\lim_{n\to\infty}A_{n}=\lim_{n\to\infty}\frac{  n}{2}\sin\bigg(\dfrac{2\pi}{n}\bigg)
    or equivalently
    \lim_{n\to\infty}\frac{n}{2\pi}\sin\bigg(\dfrac{2\  pi}{n}\bigg)=1\qquad(*).
    Now let u:=2\pi/n, and as n\to\infty, we have u\to0, therefore (*) takes the following form
    \lim_{u\to0}\frac{\sin(u)}{u}=1.
    I hope this helps. :]
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  12. #12
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    Alternatively, look at the unit circle.




    It should be pretty clear from the diagram that the area of the sector is squeezed between the area of the two triangles made by \sin{\theta} and \tan{\theta}.


    So we have

    \frac{1}{2}\sin{\theta}\cos{\theta} \leq \frac{1}{2}\theta \leq \frac{1}{2}\tan{\theta}

    \sin{\theta}\cos{\theta} \leq \theta \leq \frac{\sin{\theta}}{\cos{\theta}}

    \frac{\cos{\theta}}{\sin{\theta}} \leq \frac{1}{\theta} \leq \frac{1}{\sin{\theta}\cos{\theta}}

    \cos{\theta} \leq \frac{\sin{\theta}}{\theta} \leq \frac{1}{\cos{\theta}}.


    Now apply the sandwich theorem by letting \theta \to 0.
    Last edited by Prove It; September 20th 2009 at 01:49 AM.
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    Yes, both of those helped immensely. Thanks for all your help!
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