# Thread: Problem with a limit's existence...

1. ## Problem with a limit's existence...

Heya guys. I have to study the continuity of the following function at (0,0):

$f(x,y) =\begin{cases}
\frac{x^3y}{x^4+y^2} & (x,y)\neq(0,0) \\
0 & (x,y)=(0,0)
\end{cases}$

Thus I need to check whether

$\lim_{(x,y)\to (0,0)} \frac{x^3y}{x^4+y^2}$

exists or not.

I converted to polar coordinates:

$\lim_{r\to 0} \frac{r^4 cos(\theta)^3 sin(\theta)}{(r cos(\theta))^4+(r sin(\theta))^2} = \lim_{r\to 0} \frac{r^2 cos(\theta)^3 sin(\theta)}{r^2 cos(\theta)^4+sin(\theta)^2} = \frac{0}{sin(\theta)^2}$

Thus the limit seems to be 0 except where

$sin(\theta)^2=0 \implies sin(\theta)=0 \implies \theta = 0 \vee \theta = \pi$

If $\theta = 0 \vee \theta = \pi$, then

$\lim_{r\to 0} \frac{r^4 cos(\theta)^3 sin(\theta)}{(r cos(\theta))^4+(r sin(\theta))^2} = \frac{0}{0}$

I then used L'Hopital's rule:

$\lim_{r\to 0} \frac{r^2 cos(\theta)^3 sin(\theta)}{r^2 cos(\theta)^4+sin(\theta)^2} = \lim_{r\to 0} \frac{cos(\theta)^3 sin(\theta)}{cos(\theta)^4} = 0$

since $sin(\theta)=0$.

However, by using such paths as $y=x$ and $y=2x$, the limit equals 1 and 2, respectively. What's going on that I don't understand here?

Thanks!

2. Originally Posted by FatherChaos
I have to study the continuity of the following function at (0,0):

$f(x,y) =\begin{cases}
\frac{x^3y}{x^4+y^2} & (x,y)\neq(0,0) \\
0 & (x,y)=(0,0)
\end{cases}$

... [snip] ...

However, by using such paths as $\color{magenta}y=x$ and $\color{magenta}y=2x$, the limit equals 1 and 2, respectively.
That last statement is not true. If y=kx (with k ≠ 0) then $f(x,y) = \frac{kx^4}{x^4+y^2} = \frac{kx^2}{x^2+k^2}$, and so $\lim_{(x,y)\to(0,0)}f(x,y) = 0$.

Hint: write f(x,y) as $x\cdot\frac{x^2y}{x^4+y^2}$, and show that $\frac{x^2y}{x^4+y^2}\leqslant\frac12$.

3. Also see the following graphic.

..........The numerator, Absolute value of the numerator, The denominator

Codes for Mathematica

Code:
Plot3D[{Abs[x^3y],x^3y,x^4+y^2},{x,-1/2,1/2},{y,-1/2,1/2},PlotStyle->{{Red},{Yellow},{Blue}}]
Can you see that the (absolute value of) numerator is closer to $0$ than the denominator?
So this implies that the limit must be $0$ at $(0,0)$.
To prove this mathematically, you can lead the direction Opalg points.

4. Originally Posted by Opalg
That last statement is not true. If y=kx (with k ≠ 0) then $f(x,y) = \frac{kx^4}{x^4+y^2} = \frac{kx^2}{x^2+k^2}$, and so $\lim_{(x,y)\to(0,0)}f(x,y) = 0$.

Hint: write f(x,y) as $x\cdot\frac{x^2y}{x^4+y^2}$, and show that $\frac{x^2y}{x^4+y^2}\leqslant\frac12$.
Thanks a lot. Somehow my teacher's corrected version of this exercise was flat-out wrong. I still have a question though. What is the point of showing that $\frac{x^2y}{x^4+y^2}\leqslant\frac12$, exactly? I do not get the big picture.

5. Originally Posted by FatherChaos
What is the point of showing that $\frac{x^2y}{x^4+y^2}\leqslant\frac12$, exactly? I do not get the big picture.
The point is that it then follows that $|f(x,y)|= \left|x\,\frac{x^2y}{x^4+y^2}\right|\leqslant\tfra c12|x|$, and so $\lim_{(x,y)\to(0,0)}f(x,y) = 0.$

6. Originally Posted by Opalg
The point is that it then follows that $|f(x,y)|= \left|x\,\frac{x^2y}{x^4+y^2}\right|\leqslant\tfra c12|x|$, and so $\lim_{(x,y)\to(0,0)}f(x,y) = 0.$
Wow, you're right! And I came to the conclusion that $\frac{x^2y}{x^4+y^2}\leqslant\tfrac12 \implies (x^2-y)^2\geqslant 0$, which is obviously true. Thus, the limit exists. Thanks a bunch ! Also, one last question. Why is it that I always see those absolute values when you use the squeeze theorem? Are they really needed?