# Math Help - Limit of Vector Valued Function

1. ## Limit of Vector Valued Function

How would I approach a question such as:
find the limits:
1-lim{ (x^4 + y^4) / (x^2+ y^2)} as (x,y) tends to (0,0)
2-lim{ xy^2 / (x^2 + y^4)} as (x,y) tends to (0,0)

Thanks

2. edit: nvm what i just said.

polar coordinate that Twig showed is the way to go. It drops you down from two variables (x, y) to one (r).

3. Hi

1st one:

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

Use polar coordinates, $x = r cos(\theta) \, \, y = r sin(\theta)$

Gives us: $\frac{r^4(cos^{4}(\theta) + sin^{4}(\theta))}{r^2} = r^2(cos^{4}(\theta) + sin^{4}(\theta))$

And here $|(cos^{4}(\theta) + sin^{4}(\theta))| \leq 2 \, \Rightarrow r^2(cos^{4}(\theta) + sin^{4}(\theta)) \mbox{ goes to zero as r approaches zero}$

2nd one:

If we let $x = 0$, then the expression is already zero, which approaches zero.

But if we let $x = y^2 \, \Rightarrow \frac{y^4}{2y^4} \, \mbox{ which approaches } \frac{1}{2}$

What does this tell you?

4. Originally Posted by Twig
Hi

1st one:

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

Use polar coordinates, $x = r cos(\theta) \, \, y = r sin(\theta)$

Gives us: $\frac{r^4(cos^{4}(\theta) + sin^{4}(\theta))}{r^2} = r^2(cos^{4}(\theta) + sin^{4}(\theta))$

And here $|(cos^{4}(\theta) + sin^{4}(\theta))| \leq 2 \, \Rightarrow r^2(cos^{4}(\theta) + sin^{4}(\theta)) \mbox{ goes to zero as r approaches zero}$

2nd one:

If we let $x = 0$, then the expression is already zero, which approaches zero.

But if we let $x = y^2 \, \Rightarrow \frac{y^4}{2y^4} \, \mbox{ which approaches } \frac{1}{2}$

What does this tell you?
Hi
It tells me there is no limit at the origin since there are points close to the origin which the function takes the value 1/2 and not 0 ?

Why use polar coordinates on the first one?
Also, are there any pointers you can advise on how to approach these questions, i.e. finding the value of a limit at a point(for vector valued fns)?

5. hi

I used polar coordinates because I find it convenient and relatively easy here to show that no matter what angle we approach by, the function tends to zero.
I´m not that familiar with vector valued functions to be honest, I have only done these types of limits when $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ .

But I would imagnine that both limits have to exist for a vector to approach some point so to speak.