Two variables limit

Homework Statement

$$\lim_{(x,y) \rightarrow (0,2)} \dfrac{ysinx}{x}$$

The Attempt at a Solution

I know that I have to evaluate the function in the given values of x and y

For y=2

$$\lim_{x \rightarrow 0} \dfrac{2sinx}{x}$$

Using L'Hopital

$$\lim_{x \rightarrow 0} \dfrac{2cosx}{1}=2$$

For x=0

$$\lim_{y \rightarrow 2} \dfrac{ysin0}{0}$$

I don't know how to solve that. Thanks for your time.

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LCKurtz
Homework Helper
Gold Member
But you can't just plug in (0,2) because the function isn't defined there. And if you think the two variable limit exists, you can't prove it by trying different paths.

Since you know that sin(x)/x → 1 as x → 0 and you know y → 2 can you figure out how to use one of the limit theorems here?

HallsofIvy
Homework Helper

Homework Statement

$$\lim_{(x,y) \rightarrow (0,2)} \dfrac{ysinx}{x}$$

The Attempt at a Solution

I know that I have to evaluate the function in the given values of x and y
No, you don't. Where did you get that idea? What you often can do is show that taking the limit from two different ways gives two different results, proving that the limit does not exist but they don't have to be "horizontal" and "vertical". And, of course, that won't prove a limit does exist.

For y=2

$$\lim_{x \rightarrow 0} \dfrac{2sinx}{x}$$

Using L'Hopital

$$\lim_{x \rightarrow 0} \dfrac{2cosx}{1}=2$$

For x=0

$$\lim_{y \rightarrow 2} \dfrac{ysin0}{0}$$

I don't know how to solve that. Thanks for your time.

But you can't just plug in (0,2) because the function isn't defined there. And if you think the two variable limit exists, you can't prove it by trying different paths.

Since you know that sin(x)/x → 1 as x → 0 and you know y → 2 can you figure out how to use one of the limit theorems here?
So If I put the limit as

$$\lim_{x \rightarrow 0}\lim_{y \rightarrow 2} \dfrac{ysinx}{x}=\lim_{x \rightarrow 0}\dfrac{2sinx}{x}=2$$

Since $$\lim_{y \rightarrow 2} \dfrac{ysinx}{x}=\dfrac{2sinx}{x}$$

Am I right?

LCKurtz
Homework Helper
Gold Member
So If I put the limit as

$$\lim_{x \rightarrow 0}\lim_{y \rightarrow 2} \dfrac{ysinx}{x}=\lim_{x \rightarrow 0}\dfrac{2sinx}{x}=2$$

Since $$\lim_{y \rightarrow 2} \dfrac{ysinx}{x}=\dfrac{2sinx}{x}$$

Am I right?
That doesn't prove the limit is 2 along any path, just that one path.

That doesn't prove the limit is 2 along any path, just that one path.
So I only proved it approaching by two rects, if I approach the limit by paraboloids, for example

For $$y=x^2$$

$$\lim_{x \rightarrow 0} \dfrac{x^2sinx}{x}=\lim_{x \rightarrow 0} {xsinx}=0$$

Then I get two different values (the limit is 2 by the rects and is 0 by a paraboloid), so I don't know if now I can say the limit does not exist given that the limit evaluated by two different paths is not the same.

HallsofIvy
No, you can't approach the point (0,2) on the curve $y= x^2$! (0, 2) is not on that curve.
$$\left | \frac {y\sin(x)}{x}-2\right|$$