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Thread: Dealing with Multivariate Limits?

  1. #1
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    Dealing with Multivariate Limits?

    Just revising, and came across a few limits I couldn't remember how to prove.
    1. lim(x,y)-->(0,0) (cosx-1+x^2/2)/(x^4-y^4)
    2. lim(x,y)-->(0,0) (y-2x+sin2x)/(x^3+y)

    And these, which are supposed to use sandwich rule
    1. 7x^2*y^2/(x^2+2y^4)
    2. 3y*x^2/(x^2+y^2)


    Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    The most promising way to find such types of limits is working in polar coordinates with the substitution...

    x= \rho \cdot \cos \theta

    y = \rho \cdot \sin \theta (1)

    ... and searching [if it exists...] the limit with \rho \rightarrow 0 that is independent from \theta.

    Let's start with your example nr. 2. With substitutions (1) we have...

    \lim_{[x,y]\rightarrow [0,0]} \frac{y- 2\cdot x + \sin x}{x^{3}+y} = \lim_{\rho \rightarrow 0} \frac {\rho\cdot (\sin \theta - 2\cdot \cos \theta) + \sin (2\cdot \rho \cdot \cos \theta)}{\rho \cdot (\rho^{2}\cdot \cos^{3}\theta+ \sin \theta)} (2)

    For 'small' value of \rho is...

    \sin (2\cdot \rho \cdot \cos \theta) \approx 2\cdot \rho \cdot \cos \theta

    \rho^{2}\cdot \cos^{3}\theta+ \sin \theta \approx \sin \theta (3)

    ... so that...

    \lim_{\rho \rightarrow 0} \frac {\rho\cdot (\sin \theta - 2\cdot \cos \theta) + \sin (2\cdot \rho \cdot \cos \theta)}{\rho \cdot (\rho^{2}\cdot \cos^{3}\theta+ \sin \theta)} = \lim_{\rho \rightarrow 0} \frac {\rho \cdot (\sin \theta - 2\cdot \cos \theta + 2\cdot \cos \theta)}{\rho \cdot \sin \theta} = 1 (4)

    And now your example nr. 1. With substitutions (1) we have...

    \lim_{[x,y]\rightarrow [0,0]} \frac{\cos x - 1 + \frac{x^{2}}{2}}{x^{4}- y^{4}} = \lim_{\rho \rightarrow 0} \frac{\cos(\rho\cdot \cos \theta) -1 + \frac{\rho^{2}\cdot \cos^{2}\theta}{2}}{\rho^{4}\cdot (\cos^{4} \theta - \sin^{4} \theta)} (5)

    For 'small' value of \rho is...

     \cos (\rho\cdot \cos \theta) \approx 1 - \frac {\rho^{2}\cdot \cos^{2} \theta}{2} + \frac {\rho^{4}\cdot \cos^{4} \theta}{24} (6)

    ... so that...

    \lim_{\rho \rightarrow 0} \frac{\cos(\rho\cdot \cos \theta) -1 + \frac{\rho^{2}\cdot \cos^{2}\theta}{2}}{\rho^{4}\cdot (\cos^{4} \theta - \sin^{4} \theta)} = \frac{1}{24}\cdot \frac{\cos^{4} \theta}{\cos^{4} \theta - \sin^{4} \theta} (7)

    Now we observe that the limit (7) does depend from \theta so that we must conclude that limit (5) doesn't exist...

    Kind regards

    \chi \sigma
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