Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The title says it all: does the following limit exist?

$$\lim_{\large(x, y) \to (1, 2)} \; 3x^3 - x^2y^2$$

It approaches $\,-1\,$ with direct substitution, but if you approach the point with the curve $x = y^2$, you get $\,128\,$ which is not $\,-1,\,$ meaning the limit doesn't exist. I got the problem wrong but I am curious to see if it does actually exist.

I'm approaching with $\,x = y^2.\,$ So it ends up being $\;\displaystyle\lim_{y\to 2}\;3y^6-y^6,\,$ which is $128$.

share|improve this question
    
do you mind showing your computation for the limit along the curve? –  Ittay Weiss Feb 12 '13 at 0:59
2  
This function is continuous on $\mathbb{R}^2$, so in particular at $(1,2)$. So... –  1015 Feb 12 '13 at 1:01
    
What proof do you have of the function being discontinuous at (1,2)? If it is continuous there then the direct substitution is one way to compute the answer. –  JB King Feb 12 '13 at 1:03
    
Our book says to check from all directions and gives an example similar to this saying that the limit approaches the same thing from the x direction and the y direction but then why computing for y^2 it gets a different answer concluding that the limit DNE –  user61918 Feb 12 '13 at 1:05
1  
If $y \to 2$ and $x=y^2$ then $x \to 4$... So the second limit you are calculating is $\lim_{(x,y) \to (4,2)}3x^3-x^2y^2$ along some particular curve.... –  N. S. Feb 12 '13 at 1:30

2 Answers 2

up vote 4 down vote accepted

$(x, y) \to (1, 2),\;$ but $\;(1, 2) \notin x=y^2$

The function is continuous on $\mathbb{R}^2,\;$ hence at $(1,2)$, so direct substitution is suffices to compute the limit, as you started out doing.

I think you're making this more complicated than necessary.

share|improve this answer

If the curve you are following along is $x=y^2$ then if $x=1$, we get $y=1$ on the curve. Likewise if $y=2$ then $x=\sqrt{2}$. You cannot approach point $(1,2)$ along the curve $x=y^2$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.