Lim x,y→0: Multivariable Limit

[nate9519]

Homework Statement

find lim as x,y approach 0 of (10sin(x^2 + y^2)) / (x^2 + y^2)

Homework Equations

The Attempt at a Solution

direct substitution yields indeterminate form and so does multiplying by the conjugate. what other methods are there to use?

 

[STEMucator]

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.

 

[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

 

[STEMucator]

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

What is:

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

Looks like a first year problem.

 

[nate9519]

wow. can't 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

 

[STEMucator]

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.

 

[nate9519]

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

 

[STEMucator]

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

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: $\displaystyle \lim_{x \rightarrow 0} \frac{sin(x)}{x} = 1$

 

[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

 

[Ray Vickson]

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

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 $f(x,y)$.

 

[vela]

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 it's like that variable never existed?

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.