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Anyone can help me with this limit?

$$\lim_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$

I'm having trouble with proving that this limit really goes to $0$

thank you

Martin Sleziak
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JP91
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4 Answers4

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Let $y=(x^3-x^6)^2-x^2$. $\,$Then \begin{align*} \frac{x^3+y^3}{x^2+y} &= \frac{x^3+(x^3-x^6)^6 - x^6 + 3x^4(x^3-x^6)^2 - 3 x^2 (x^3-x^6)^4}{(x^3-x^6)^2}\\ &= \frac{1}{x^3-x^6} + (x^3-x^6)^4+3x^4-3x^2(x^3-x^6)^2\\ &\rightarrow \infty. \end{align*} That is, \begin{align*} \lim_{(x,y)\rightarrow (0, 0)} \frac{x^3+y^3}{x^2+y} \end{align*} does not exist.

Gordon
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  • If $x>0$ the limit goes to $-\infty$ but if $x<0$ the limit goes to $\infty$ so the 2 sided limit doesn't exist – kingW3 Sep 18 '15 at 20:36
  • @kingW3, For a path, the limit is $\infty$, while for some other paths the limits are $0$ (easy to find such paths, e.g., $y=kx$). That is, the limits from different paths are different. Then the limit does not exist. – Gordon Sep 18 '15 at 21:54
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Hint: use $x=r\cos\Theta‎$ and $y=r\sin\Theta‎$ you will see $\lim_{r\to 0}\frac{(r\cos\Theta)^3+(r\sin\Theta)^3}{(r\cos\Theta)^2+(r\sin\Theta)}=\lim_{\to 0}\frac{r^3((\cos\Theta)^3+(\sin\Theta)^3)}{r(r\cos^2\Theta+\sin\Theta)}=\lim_{\to 0}\frac{r^2((\cos\Theta)^3+(\sin\Theta)^3)}{(r\cos^2\Theta+\sin\Theta)}$ it depends on $\Theta$ thus limit doesnot exist

R.N
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  • you do that substituition just to make the numerator power higher than the denominator power, and then prove that the limit is 0? – JP91 Sep 18 '15 at 02:36
  • yes, you are right – R.N Sep 18 '15 at 05:34
  • @RaziehNoori,How does it prove the existence of the limit??Actually what you have done is find a limit along a line( through origin and at an angle $\theta$ from +ve x axis). – Koro Sep 18 '15 at 14:13
  • you are right it may have pro with $(r\cos^2\Theta+\sin\Theta)$ because of power of $sin\Theta$. thank you dear – R.N Sep 18 '15 at 15:08
  • @RaziehNoori,Can we actually find out a curve like say $y=g(x)$ passing through origin which makes the value of the limit $\ne0$?? – Koro Sep 18 '15 at 15:13
  • oh my god, how much a question that i thought is easy, have been hard? let me think more. if i get a result i will say. please if you get do same(sorry for weak English) – R.N Sep 18 '15 at 15:24
  • what about $y=-x^2$ @kilimanjaro – R.N Sep 18 '15 at 15:29
  • No,that can't work. – Koro Sep 18 '15 at 15:31
  • because of domain :( – R.N Sep 18 '15 at 15:35
  • @RaziehNoori,yes,to a good extent ! – Koro Sep 18 '15 at 15:56
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    Wolfram says the limit exists http://www.wolframalpha.com/input/?i=lim_%28%7Bx%2C+y%7D-%3E%7B0%2C+0%7D%29%28x%5E3%2By%5E3%29%2F%28x%5E2%2By%29 – kingW3 Sep 18 '15 at 16:37
  • @kingW3, Wolfram also states that this limit is also 0, which is obviously false if you take the paths $y=x$ and $y=x^2$. – k170 Sep 18 '15 at 18:41
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$$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}$$ Method 1: Using the two path test

Along the path of $y=0$, we have $$\lim\limits_{(x,0)\to (0,0)} \frac{x^3 + 0^3}{x^2 + 0}=\lim\limits_{x\to 0} \frac{x^3}{x^2}=\lim\limits_{x\to 0} x=0$$ Along the path of $y=p(x)-x^2$, where $p(x)$ is a polynomial that passes through the origin, we have $$\lim\limits_{(x,p(x)-x^2)\to (0,0)} \frac{x^3 + (p(x)-x^2)^3}{x^2 + p(x)-x^2}$$ $$=\lim\limits_{x\to 0} \frac{x^3 -x^6+ 3x^4p(x)-3x^2p(x)^2+p(x)^3}{p(x)}$$ $$=\lim\limits_{x\to 0} \left[\frac{x^3 -x^6}{p(x)}+ 3x^4-3x^2p(x)+p(x)^2\right]$$ Let $p(x)=x^3-x^6$, then $$\lim\limits_{x\to 0} \left[1+ 3x^4-3x^2(x^3-x^6)+(x^3-x^6)^2\right]=1$$ Since we have different values along different paths, we can conclude that $$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}=\mbox{non existent}$$ Method 2: Using polar coordinates

$$\lim\limits_{r\to 0^+} \frac{r^3\cos^3\phi + r^3\sin^3\phi}{r^2\cos^2\phi + r\sin\phi}$$ $$=\lim\limits_{r\to 0^+} r^2\left(\frac{\cos^3\phi + \sin^3\phi}{r\cos^2\phi + \sin\phi}\right)$$ Now lets attempt to find bounds that are independent of $\phi$. Since $$\left|\cos^3\phi + \sin^3\phi\right|\leq 1$$ We have $$r^2\left|\frac{\cos^3\phi + \sin^3\phi}{r\cos^2\phi + \sin\phi}\right|\leq \frac{r^2}{\left|r\cos^2\phi + \sin\phi\right|}$$ Note that $\phi$ is a variable and we cannot treat it as a constant. Since the right hand side is dependent on $\phi$, then we can conclude that $$\lim\limits_{(x,y)\to (0,0)} \frac{x^3 + y^3}{x^2 + y}=\mbox{non existent}$$

k170
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  • @k170,you are welcome.But please help me know :can we apply sandwich theorem in Polar coordinate? – Koro Sep 18 '15 at 14:26
  • @kilimanjaro, yes we can use the squeeze theorem however care must be taken when evaluating the bounding limits. In other words, we must find a way to remove the dependency on $\phi$. In this case, we cannot, hence the limit does not exist. – k170 Sep 18 '15 at 19:42
  • @kilimanjaro, I've updated my answer to include a method of finding an alternate path. – k170 Sep 25 '15 at 14:01
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Putting polar coordinates $x=r\sin \theta,y=r\cos \theta$ $$\lim_{r\to 0} \frac{r^3\sin^3\theta+r^3\cos^3\theta}{r^2\sin^2 \theta + r\cos \theta}=\lim_{r\to 0}\frac{r(\sin \theta + \cos \theta)}{\sin^2\theta+\frac{\cos \theta}{r}}$$ Now if $\cos \theta \not\to 0$ then $\frac{|\cos \theta|}{|r|}\to \infty$ hence the limit is $0$ if $\cos\theta \to 0$ then we have that $|\sin \theta|\to 1$ and since denominator tends to $1$ and numerator tends to $0$ again the limit is $0$ hence the limit is $0$ independent of $\theta$

kingW3
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