# Why does limit $\lim_{(x,y) \to (0,0)}\frac{2x}{x^2+x+y^2}$ not exist?

I don't understand how this limit doesn't exist.

\begin{align*} \lim_{(x,y) \to (0,0)}\dfrac{2x}{x^2+x+y^2}&= \lim_{r \to 0}\dfrac{2r\cos\theta}{r^2\cos^2\theta+r\cos\theta+r^2\sin^2\theta}\\ &=\lim_{r \to 0}\dfrac{2r\cos\theta}{r^2+r\cos\theta}\\ &=\lim_{r \to 0}\dfrac{2\cos\theta}{r+\cos\theta} \\ &= 2 \end{align*}

I find that limit goes to $2$. But by WolframAlpha, limit does not exist. How does that work ?

• According to Wolfram, it is due to the fact that $x$ and $y$ may take on different values in the complex plane. – Ryan Goulden Mar 29 '17 at 7:45
• Problem comes when you try to simplify $\cos\theta / \cos\theta$. You must make sure that $\cos\theta \neq 0$. – Zubzub Mar 29 '17 at 13:58

The other answers give excellent reasons as to why the "limit does not exist" but here is more of a reason "why" your logic breaks down at the last line (which answers your question "How does that work?"):

In the last line we should really write

$$\lim_{r\to0}\dfrac{2\cos\theta}{r+\cos\theta}=\dfrac{2\cos\theta}{\cos\theta}=2 \text{, if }\cos\theta\neq0.$$

Because in that last step, by 'cancelling' $\cos\theta$ from the numerator and denominator, you want to make sure that you're not dividing by zero. Since $\cos\theta=0$ is a definite possibility, then we need to consider it as a separate case.

As suggested by @Patrick, we can get this scenario by setting $\theta=\tfrac{\pi}2$, yielding the sub-case

$$\left.\lim_{r\to0}\dfrac{2\cos\theta}{r+\cos\theta}\right|_{\theta=\tfrac{\pi}2}=\lim_{r\to0}\dfrac{0}{r+0}=0,$$

hence showing that we can get a different answer to the limit, hence it does not exist.

The limit doesn't exist. Fix $y=0$; then the limit expression is $$\lim_{x \to 0} \frac{2}{x+1} = 2$$

Fix $x=0$; then the limit expression is identically $$\lim_{y \to 0} \frac{0}{y^2} = 0$$

• But our instructor said that we use polar coordinate – K. Talha Mar 29 '17 at 8:04
• In that case, set $r=0$ and $\theta = \frac{\pi}{2}$ in the two cases respectively. It doesn't make a difference. – Patrick Stevens Mar 29 '17 at 8:10

You neglected to control $\theta$ too.

• The answer should still be 2, given that $/theta$ and $r$ are real though, no? – Ryan Goulden Mar 29 '17 at 7:52
• This should be a comment and not an answer – lioness99a Mar 29 '17 at 8:14