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I'm trying to compute the following limit:

$\lim_{(x,y)\to(0,0)}x\cos\bigg(\frac{2\pi x^2}{x^2 +y^2}\bigg)$

I checked wolfram and it said the limit is zero. However I don't understand why because if we isolate $\cos\bigg(\frac{2\pi x^2}{x^2 +y^2}\bigg)$ this limit doesn't exist (because the direccional limits are different). Maybe I'm not applying correct properties.

Also how do I justify the limit exist. Can I just say it's zero because it's the product of a function that ends to zero by a limited function (cosine). How is there any majoration (like we usually do in multivariable calculus).

I just need a bit of clarification. Thanks!

Granger Obliviate
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    Use the squeeze theorem. Away from $(0,0)$ the cosine part is bounded by $1$. – Umberto P. Jan 27 '17 at 16:47
  • The limit of the cosine does not exist, but the cosine is bounded. So when you multiply by $x$, the product tends to zero anyway. –  Jan 27 '17 at 17:06

2 Answers2

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Hint: No matter what is inside the cosine (as long as it is real valued) the cosine will always be bounded in absolute value by $1$. Now apply the squeeze theorem

b00n heT
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Another way is by using polar coordinates. $$\lim _{ (x,y)\to (0,0) } x\cos \left( \frac { 2\pi x^{ 2 } }{ x^{ 2 }+y^{ 2 } } \right) =\lim _{ r\rightarrow { 0 }^{ + } }{ r\cos { \theta \cdot \cos { \left( \frac { 2\pi { r }^{ 2 }\cos ^{ 2 }{ \theta } }{ { r }^{ 2 } } \right) } } } =\lim _{ r\rightarrow { 0 }^{ + } }{ r\cos { \theta \cdot \cos { \left( 2\pi \cos ^{ 2 }{ \theta } \right) } } } =0$$

haqnatural
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