Two Variables Limit: Homework Solution

[krozer]

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.

 

[LCKurtz]

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 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.

 

[krozer]

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]

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.

 

[krozer]

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.

 

[LCKurtz]

Both Halls and I have been telling you that you can't show the limit exists by checking different paths. You haven't addressed my suggestion in post #2 of using the limit theorems. Alternatively you can show the difference between the function and 2 can be made small by estimating

\left | \frac {y\sin(x)}{x}-2\right|
If you try this you might start by adding and subtracting y and grouping terms appropriately.

Personally, I would use the limit theorems...