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partial derivatives

Here's the question I attempted and the worked solution for it in blue:

https://cdn.discordapp.com/attachments/692344148212187226/1009838232650928288/Screenshot_46.png

Here's my working:

https://media.discordapp.net/attachments/692344148212187226/1009838807153127574/20220818_155817.jpg?width=1025&height=404

The issue I have is that I followed a method used in an example I looked at. The method involved evaluating the limit of the function via taking either the x or y-axis path and seeing if the end result was equal to the number below the main function within the curly bracket, if it was equal then the function would be continuous at (0,0). I did this and my end result was 0 which was equal to the value at (0,0) yet the mark scheme is saying it's not continuous at that point. Which method is correct?
Original post by Soul Wavel3ngth
Here's the question I attempted and the worked solution for it in blue:

https://cdn.discordapp.com/attachments/692344148212187226/1009838232650928288/Screenshot_46.png

Here's my working:

https://media.discordapp.net/attachments/692344148212187226/1009838807153127574/20220818_155817.jpg?width=1025&height=404

The issue I have is that I followed a method used in an example I looked at. The method involved evaluating the limit of the function via taking either the x or y-axis path and seeing if the end result was equal to the number below the main function within the curly bracket, if it was equal then the function would be continuous at (0,0). I did this and my end result was 0 which was equal to the value at (0,0) yet the mark scheme is saying it's not continuous at that point. Which method is correct?


The markscheme is correct. If the limit as (x,y) goes to (0,0) doesn't exist, then there's no way it can be continuous at that point.

I suggest posting the example you looked at (with working) whch contains the method you've used.
Original post by Soul Wavel3ngth
Here's the question I attempted and the worked solution for it in blue:

https://cdn.discordapp.com/attachments/692344148212187226/1009838232650928288/Screenshot_46.png

Here's my working:

https://media.discordapp.net/attachments/692344148212187226/1009838807153127574/20220818_155817.jpg?width=1025&height=404

The issue I have is that I followed a method used in an example I looked at. The method involved evaluating the limit of the function via taking either the x or y-axis path and seeing if the end result was equal to the number below the main function within the curly bracket, if it was equal then the function would be continuous at (0,0). I did this and my end result was 0 which was equal to the value at (0,0) yet the mark scheme is saying it's not continuous at that point. Which method is correct?


This is a good example of a 0 limit along the x-axis (when y=0) but what we mean when we write

lim(x,y)(0,0)f(x,y)=0\displaystyle \lim_{(x,y) \to (0,0)} f(x,y) = 0

is that the limit is zero along ALL possible paths that (x,y)(x,y) take to approach (0,0)(0,0).

You found one path on which it works, but now consider the path y=xy = x on which we get

lim(x,y)(0,0)xyx2+y2=limx0x2x2+x2=12\displaystyle \lim_{(x,y) \to (0,0)} \dfrac{xy}{x^2 + y^2} = \lim_{x \to 0} \dfrac{x^2}{x^2 + x^2} = \dfrac{1}{2}

which is clearly not 0.

So the limit of the function is not 0. It does not exist!
(edited 1 year ago)
Original post by ghostwalker
The markscheme is correct. If the limit as (x,y) goes to (0,0) doesn't exist, then there's no way it can be continuous at that point.

I suggest posting the example you looked at (with working) whch contains the method you've used.


My bad I forgot to link it but here (it's segmented):

https://media.discordapp.net/attachments/692344148212187226/1009839892676751491/20220818_160058.jpg?width=1025&height=329


https://media.discordapp.net/attachments/692344148212187226/1009839892924203141/20220818_160250.jpg?width=1025&height=467
This clears things up. Thanks a lot.

Original post by RDKGames
This is a good example of a 0 limit along the x-axis (when y=0) but what we mean when we write

lim(x,y)(0,0)f(x,y)=0\displaystyle \lim_{(x,y) \to (0,0)} f(x,y) = 0

is that the limit is zero along ALL possible paths that (x,y)(x,y) take to approach (0,0)(0,0).

You found one path on which it works, but now consider the path y=xy = x on which we get

lim(x,y)(0,0)xyx2+y2=limx0x2x2+x2=12\displaystyle \lim_{(x,y) \to (0,0)} \dfrac{xy}{x^2 + y^2} = \lim_{x \to 0} \dfrac{x^2}{x^2 + x^2} = \dfrac{1}{2}

which is clearly not 0.

So the limit of the function is not 0. It does not exist!


:holmes:

Well in this example the function is continuous (x,y)R2\forall (x,y)\in\mathbb{R}^2, but the argument given for it being continuous at (0,0) is insufficient.

Would be better to use a power series expansion, which simplifies quite nicely.

Is this from a textbook? Do you have an image of the original.
Original post by ghostwalker
:holmes:

Well in this example the function is continuous (x,y)R2\forall (x,y)\in\mathbb{R}^2, but the argument given for it being continuous at (0,0) is insufficient.

Would be better to use a power series expansion, which simplifies quite nicely.

Is this from a textbook? Do you have an image of the original.


There is a trap here when using anything like Taylor series.

Setting (x,y)=(rcosθ,rsinθ)(x,y)=(\sqrt{r}\cos\theta,\sqrt{r}\sin\theta) turns this limit into

lim(x,y)(0,0)ex2+y21x2+y2=limr0er1r\displaystyle \lim_{(x,y)\to(0,0)} \dfrac{e^{x^2+y^2}-1}{x^2+y^2} = \lim_{r\to 0} \dfrac{e^r-1}{r}

This limit is by definition the derivative of exe^x at x=0 yet by using Taylor power series expansion we would be assuming we already know the limit is 1. Circular logic.

We need to use the definition of e however it was presented to OP, so this is a tricky question.

https://math.stackexchange.com/questions/2832934/how-to-prove-that-lim-h-to-0-fraceh-1h
(edited 1 year ago)

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