Homework Help: Multivariable limit

1. Oct 30, 2014

nate9519

1. The problem statement, all variables and given/known data
find lim as x,y approach 0 of (10sin(x^2 + y^2)) / (x^2 + y^2)

2. Relevant equations

3. The attempt at a solution
direct substitution yields indeterminate form and so does multiplying by the conjugate. what other methods are there to use?

2. Oct 30, 2014

Zondrina

So you want to find:

$$\displaystyle \lim_{(x,y) \rightarrow (0,0)} \frac{10sin(x^2 + y^2)}{x^2 + y^2}$$

Try approaching $(0, 0)$ along different paths to zero such as $(0, x)$ and $(x, 0)$.

EDIT: Path $(t, t)$ looks promising.

3. Oct 30, 2014

nate9519

for the paths (0,x) and (x,0) I got 10sin(y^2) / y^2. does that mean the limit is 10sin(y^2) / y^2

4. Oct 30, 2014

Zondrina

What is:

$$\displaystyle \lim_{x \rightarrow 0} \frac{10sin(x^2)}{x^2}$$

Looks like a first year problem.

5. Oct 30, 2014

nate9519

wow. cant believe I didn't see that . so that is indeterminate but I know the limit exists because the problem says "Hint - the limit does exist". when you said the path (t,t) looked promising I assumed you meant parameterizing x and y. but what do I let them equal

6. Oct 30, 2014

Zondrina

Indeed you can approach $(0, 0)$ along several different paths and get the same answer. That's how you know the limit exists and is finite. Plugging in $x = t$ and $y = t$ will give the same limit.

7. Oct 30, 2014

nate9519

so the paths (x,0) (0,x) (y,0) (0,y) and (t,t) all give indeterminate forms. Im just not seeing a way around this

8. Oct 30, 2014

Zondrina

Why not apply L'Hospital's rule? If the form is $0/0$ you can easily find the limit that way.

Although the conventional way would be to recognize: $\displaystyle \lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1$

9. Oct 30, 2014

nate9519

I was told l'hospital's rule did not apply in three dimensions. but since one variable goes away when evaluated on (x,0) , (y,0), etc... does that mean its like that variable never existed

10. Oct 30, 2014

Ray Vickson

No matter how $(x,y) \to (0,0)$ the distance of $(x,y)$ from the origin goes to 0. In other words, the squared distance $r^2 = x^2 + y^2 \to 0$. In fact, $r^2 \to 0$ if, and only if $(x,y) \to (0,0)$ in some way. Now go back and re-examine your original function $f(x,y)$.

11. Oct 31, 2014

vela

Staff Emeritus
Yes. The limits you end up with using those paths only depend on one variable, so you can use the Hospital rule on them. Unfortunately, all you'll have shown is that the limit along those paths exist. You actually need to show the original limit exists for all possible paths to the origin. You want to think about what Ray said.