# Thread: Dealing with Multivariate Limits?

1. ## 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

2. The most promising way to find such types of limits is working in polar coordinates with the substitution...

$\displaystyle x= \rho \cdot \cos \theta$

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

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

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

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

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

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

... so that...

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

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

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

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

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

... so that...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$