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I hope someone can help me. The limit is:

$$\lim_{(x,y)\to(0,0)} \frac{x-y}{\sqrt{3x^2 + 4y^2}}$$ Does this limit exist or not? I tried to solve it by using $x=r\cos\theta$ and $y=r\sin\theta$, but I can't find a solution.

I'm at this point right now:

$$\lim_{r\to 0} \frac{\cos\theta-sin\theta}{\sqrt{3\cos^2\theta+4\sin^2\theta}}$$

Am I on the right track or should I try something else?

Edit: I tried, like recommended, $x=0$ $$\lim_{(0,y)\to(0,0)} \frac{-y}{\sqrt{4y^2}}=\frac{-y}{2y}=\frac{-1}{2}$$

then $y=0$ $$\lim_{(0,y)\to(0,0)} \frac{x}{\sqrt{3x^2}}=\frac{x}{\sqrt3\sqrt{x^2}}=\frac{-1}{\sqrt{3}}$$

So it doesn't exist. Right?

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    $\begingroup$ Welcome to MSE. You should include your own work and ideas in your questions. Doing so, people would be more willing to help you. $\endgroup$ – Edu Jun 5 '18 at 12:03
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Hint:

Choose two routes: first, $\;x=0\;$ , and then $\;y=0\;$ . What did you get? Then...

In fact, only one route is enough...but whatever.

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  • $\begingroup$ Thank you for the answer. I will try it! $\endgroup$ – G. George Jun 5 '18 at 12:15
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You found by the variables substitution

$$\lim_{r\to 0} \frac{\cos\theta-\sin\theta}{\sqrt{3\cos^2\theta+4\sin^2\theta}} $$

This result show us that the limit value depends on the direction taken so the limit does not exists.

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Let $y=mx$ then $(x-y)/(3x^2+4y^2)^{1/2}$ simplifies to $(1-m)/(3+4m^2)^{1/2}$ so the direction of approach to (0,0) gives different values depending on $m$. Thus the limit does not exist.

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