Find all points that are continuous in the function

[quietrain]

1. Homework Statement
find all points that are continuous in the function f

f(x,y) = (y-5)cos(1/x2) if x not = 0
if x = 0, then f(x,y) = 0

3. The Attempt at a Solution

my notes says that to show continuity, i must show that f(x,y) = f(a,b) when x,y tends to a,b
how do i do that?
does it mean i do something like this

-1< cos(1/x2) <1
-(y-5) < (y-5)cos( 1/x2) < (y-5)
so all points are continuous from -(y-5) to (y-5) except at x = 0 ?

so for f(x,y) = 0 when x = 0, it is not continuous right? since cos(1/0) = undefined?

so all points are continuous from -(y-5) to (y-5) except at x = 0 ?

thanks!

 

[betel]

Usually the question should be stated the other way round.
Find the points (a,b) at which the function is continuous.
And as you correctly said, this is the case if the limit of the function when approaching this point is the value of the function at this point.
The problem here is that you have a two dimensional problem, so the way to approach a given point (a,b) is not unique.
You should get an answer, that the function is continuous everywhere except at (x,y)= ( certain points).

 

[SammyS]

What is the behavior of f(x,y) in the neighborhood of (x,y)=(0, 5) ?

 

[SammyS]

1. Homework Statement
find all points that are continuous in the function f

f(x,y) = (y-5)cos(1/x2), if x ≠ 0

f(x,y) = 0, if x = 0

3. The Attempt at a Solution

my notes say that to show continuity, i must show that f(x,y) = f(a,b) when x,y tends to a,b
how do i do that?
does it mean i do something like this

-1< cos(1/x2) <1
-(y-5) < (y-5)cos( 1/x2) < (y-5)
so all points are continuous from -(y-5) to (y-5) except at x = 0 ?

so for f(x,y) = 0 when x = 0, it is not continuous right? since cos(1/0) = undefined?

so all points are continuous from -(y-5) to (y-5) except at x = 0 ?

The following statement doesn't make sense:

so all points are continuous from -(y-5) to (y-5) except at x = 0 ?

What is true is that for any fixed value of y, f(x, y) takes on values from -(y-5) to (y-5), and there is a discontinuity at x=0, except if y=5.

Do your notes say how to show that lim_{(x,y)→(a,b)} f(x,y) = f(a,b) ?

 

[quietrain]

erm, they had an example where they found the limit of a function to be 0 as x,y tend to 0. so they conclude that since the function is also = 0 when x,y = 0 , which shows continuity from the theorem?
for the rest of the x,y, they concluded the function was a rational function and so it is continuous everywhere except at 0,0

so in this case, am i suppose to find the limit of (y-5)cos(1/x) for the continuity at 0,0 , but in this case, i have 2 variables...

in that example they showed, they had 2 variables too, but they use squeeze theorem to get rid of y and left with a simple upper bound limit f(x).

am i suppose to use squeeze theorem to say that (y-5)cos(1/x²) <or= (y-5)
since upper limit of cos(1/x) = 1

so i suppose i am to let y tend to 0 ? so f(y) tends to -5

but since f(0,0) is undefined due to cos1/x²
, does that mean that it is discontinuous at the point 0,0 and continuous everywhere else? since f(0,0) = undefined which is not = -5 when x,y tend to 0,0

PS: i had a typo in the question in my first post, it is cos(1/x² )

 

[quietrain]

i went to this 2 variable graphing calculator webiste
http://www.calculator-grapher.com/graphers/function-grapher-2-var.html

and i entered (y-5)(cos(1/(x^2)))
and i got a crazy graph that seems continuous everywherE?

 

[SammyS]

i went to this 2 variable graphing calculator website
http://www.calculator-grapher.com/graphers/function-grapher-2-var.html

and i entered (y-5)(cos(1/(x^2)))
and i got a crazy graph that seems continuous everywhere?

The graph IS crazy. No, it's not continuous everywhere.

erm, they had an example where they found the limit of a function to be 0 as x,y tend to 0. so they conclude that since the function is also = 0 when x,y = 0 , which shows continuity from the theorem?
for the rest of the x,y, they concluded the function was a rational function and so it is continuous everywhere except at 0,0

so in this case, am i suppose to find the limit of (y-5)cos(1/x) for the continuity at 0,0 , but in this case, i have 2 variables...

in that example they showed, they had 2 variables too, but they use squeeze theorem to get rid of y and left with a simple upper bound limit f(x).

am i suppose to use squeeze theorem to say that (y-5)cos(1/x²) <or= (y-5)
since upper limit of cos(1/x²) = 1

so i suppose i am to let y tend to 0 ? so f(y) tends to -5

but since f(0,0) is undefined due to cos1/x²
, does that mean that it is discontinuous at the point 0,0 and continuous everywhere else? since f(0,0) = undefined which is not = -5 when x,y tend to 0,0

PS: i had a typo in the question in my first post, it is cos(1/x² )

First, f(x,y) IS defined for all x & y.

It is true that cos(1/x²) is not defined for x=0, However, f(0,y) is defined as being zero. That is a BIG clue to solving this problem.

QUESTION: Do you need to prove that f(x,y) is continuous at those locations where you claim it's continuous? and prove it's not continuous at other locations? or what?

Your observation that: -(y-5) < (y-5)cos( 1/x²) < (y-5), was very good. - Look at this for any particular value of y. However, (as I have hinted at before) if y=5, then you have -0 < (y-5)cos( 1/x²) < 0, i.e. f(x,5) = 0. (More on this later.) If y≠5, then, (y-5)cos(1/x²) oscillates wildly as x→0.

Look at a 1 variable graph with y=6, i.e. Graph (6-5)cos( 1/x²)=1×cos( 1/x²) with the Grapher you found.

Look at the behavior of f(x,y) along some line passing through (0,5) the along a line not passing through (0,5)

Try y= .5 x +5. Then f(x,y)=f(x, .5 x +5) = (.5 x +5 - 5)cos( 1/x²) = (.5 x +5 - 5)cos( 1/x²).

Then try y= .5 x +4.5, which comes pretty close to (0,5). f(x, .5 x +4.5) = (.5 x - 0.5)cos( 1/x²)

You may have to mess with the scale factors to get good graphs.

 

[quietrain]

i am suppose to find all points that are continuous in the function f

f(x,y) = (y-5)cos(1/x2) if x not = 0
f(x,y) = 0 if x = 0

so from this picture

i see that for y = 5, every point of z=f(x,y) = 0.

but at points close to x, between -2 and 2, the graph is crazy, fluctuates like mad like you said :(

but it stills looks continuous at everywhere? i can't seem to find a place where there is a break

so far the conclusions are
when x=0, for any value y, the function =0
when x not = 0, for any value y except 5, the function is continous?
when x not = 0 , y = 5, the function = 0

so since x fluctuates like mad near 0 but the function f(0,y) = 0 , that means function is not continuous at x = 0 ? except at y =5 where every value of the function is 0. so continous?

btw, if i am in an exam, i can't use graphing cal, how am i suppose to know :(

 

[SammyS]

i am suppose to find all points that are continuous in the function f

f(x,y) = (y-5)cos(1/x2) if x not = 0
f(x,y) = 0 if x = 0

so from this picture [IMG ]http://i1115.photobucket.com/albums/k554/shirozack/maths1.png[/IMG [Broken] ]

i see that for y = 5, every point of z=f(x,y) = 0.

but at points close to x, between -2 and 2, the graph is crazy, fluctuates like mad like you said :(

but it stills looks continuous at everywhere? i can't seem to find a place where there is a break

so far the conclusions are
when x=0, for any value y, the function =0
when x not = 0, for any value y except 5, the function is continuous?
when x not = 0 , y = 5, the function = 0

so since x fluctuates like mad near 0 but the function f(0,y) = 0 , that means function is not continuous at x = 0 ? except at y =5 where every value of the function is 0. so continuous?

btw, if i am in an exam, i can't use graphing cal, how am i suppose to know :(

Hi quietrain.

A few comments:

That graphing site is handy. WolframAlpha is also good for some graphing.

Your graph looks nice, but all of the "action" for cos(1/x²) occurs for -1 < x < 1. In fact all of the zeros of this function occur between -0.8and 0.8 . Here's a graph of cos(1/x²) by WolframAlpha. http://usera.ImageCave.com/DGSamSnyder/cos_rec_x_x.gif.jpg The seemingly solid blue region in the center is due to the fact that the oscillations of the function get closer together as |x|→0 . The oscillations are so close together that you will never see the discontinuity at x=0 as a break in the graph. See this link.

when x ≠ 0, for any value y except 5, the function is continuous?

Correct! the function 1/x² is continuous everywhere except at x=0 . The cosine function is continuous everywhere, so the composition cos(1/x²) is continuous everywhere except at x=0.

so since x fluctuates like mad near 0 but the function f(0,y) = 0 , that means function is not continuous at x = 0 ? except at y =5 where every value of the function is 0. so continuous?

The first part is true.

The second part: ... well f(x,t) IS continuous at (x,y)=(0,0), but you need a more solid reason than that.

Example: $$g(x,y)=\left\{\begin{array}{cc}{{xy}\over{x^2+y^2}},&\mbox{ if } (x,y)\neq (0,0)\\0, & \mbox{ if } (x,y)=(0,0)\end{array}\right.$$
If x=0, or if y=0, g(x,y)=0. However, this function is not continuous at the origin. Along the line y=x, g(x,y)=g(x,x)=(x²)/(2x²)=1/2, if x≠0. Similarly, along the line y=-x, g(x,-x)=-1/2, except at the origin.
http://usera.ImageCave.com/DGSamSnyder/xy_over_r_sqared.gif.jpg See this link.
​

I'll try to leave another post tonight with ideas for proving convergence & non-convergence.

 

[quietrain]

Hi quietrain.

A few comments:

That graphing site is handy. WolframAlpha is also good for some graphing.

Your graph looks nice, but all of the "action" for cos(1/x²) occurs for -1 < x < 1. In fact all of the zeros of this function occur between -0.8and 0.8 . Here's a graph of cos(1/x²) by WolframAlpha. [PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP376319e638adi49idi3300000h14082ebc30cbe8?MSPStoreType=image/gif&s=4&w=319&h=135 [Broken] The seemingly solid blue region in the center is due to the fact that the oscillations of the function get closer together as |x|→0 . The oscillations are so close together that you will never see the discontinuity at x=0 as a break in the graph.

Correct! the function 1/x² is continuous everywhere except at x=0 . The cosine function is continuous everywhere, so the composition cos(1/x²) is continuous everywhere except at x=0.

The first part is true.

The second part: ... well f(x,t) IS continuous at (x,y)=(0,0), but you need a more solid reason than that.

Example: $$g(x,y)=\left\{\begin{array}{cc}{{xy}\over{x^2+y^2}},&\mbox{ if } (x,y)\neq (0,0)\\0, & \mbox{ if } (x,y)=(0,0)\end{array}\right.$$
If x=0, or if y=0, g(x,y)=0. However, this function is not continuous at the origin. Along the line y=x, g(x,y)=g(x,x)=(x²)/(2x²)=1/2, if x≠0. Similarly, along the line y=-x, g(x,-x)=-1/2, except at the origin.
[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP871119e62fa0ghdh7i5a00005a6d6bcai6df3c9b?MSPStoreType=image/gif&s=25&w=200&h=151 [Broken]
​

I'll try to leave another post tonight with ideas for proving convergence & non-convergence.

oh so basically, it is continuous everywhere except at x = 0
and when y = 5, everywhere is 0 but i need to find the limit of f(x,y) as it tends to (0,5) and either show the limit exist (by squeeze theorem) or show limit doesn't exist by the trial error method you showed as an example?

so if i use squeeze thereom

|(y-5)(cos1/x²)| <or= |y-5|
so y-5 is the upper bound limit
so as function tends to (0,5)
|y-5| tends to 0
so f(x,y) as (x,y)tends to (0,5) = f(0,5) = 0 = limit
so i conclude that it is continuous at 0,5? but if x = 0 is non continous, then... ?

but what about (1,5)? (2,5)? (3,5) and so on for values of x? since they all make the function = 0 which is the limit, does it mean they are all continuous too?

PS: i don't seem to be able to see your graph :X

 

[SammyS]

I fixed the links to figures in my previous post. -(I think I did ... DUH!)

BTW: Do you do δ - ε proofs?

 

[SammyS]

Discontinuity:
If x=0 AND y ≠ 5:

What we'll look at: for any given y, y≠5.

cos(n·π)=(‒1)ⁿ, so consider the sequence {x_n}, where x_n=1/(√(n·π)). This gives 1/(x_n²)=n·π, so that cos(1/x_n)=(‒1)ⁿ.

Pick ε = |y-5|/2 . No matter how small δ is,, there exists some natural number, N, such that 3N > 1/δ². Thus, if n > N, then 1/(x_n²) = n·π > 3N > 1/δ², so that if n > N, then 1/(x_n²) > 1/δ², thus x_n² < δ², so that ‒δ < x_n < δ. Now, since cos(n·π)=(‒1)ⁿ, we have that

|(y‒5)cos(n·π) ‒ 0| = |(y‒5)|·|cos(n·π)| = |(y‒5)|·|(‒1)ⁿ| = |(y‒5)| > |y-5|/2 = ε

So lim_{(x,y)→(0,y)} ≠ 0, if y ≠ 5.
​

Continuity? next post.

 

[quietrain]

my notes got say about delta and epison stuff, but i have no idea what it is talking about. i have to go look at it again

so basically the function is continuous everywhere except when x=0 and y is not = 5.

to prove it i cannot use squeeze theorem because it only shows limit exist?

i need to use the delta epsilon stuff?

:(

 

[VietDao29]

my notes got say about delta and epison stuff, but i have no idea what it is talking about. i have to go look at it again

so basically the function is continuous everywhere except when x=0 and y is not = 5.

to prove it i cannot use squeeze theorem because it only shows limit exist?

i need to use the delta epsilon stuff?

:(

You can use delta-epsilon proof. However, there's a more convenient way.

The function f is said to be continuous at (x₀, y₀) if and only if any sequence (x_n, y_n) → (x₀, y₀), we have $$\lim f (x_n, y_n) = f(x_0, y_0)$$.

So, to prove that a function f is discontinuous at some (x₀, y₀) you can choose 2 sequences (x_n, y_n), and (x'_n, y'_n), such that both of them tend to (x₀, y₀). But $$\lim f(x_n, y_n) \neq \lim f(x'_n, y'_n)$$.

Do you remember the proof that the function:
$$f:= \left\{ \begin{array}{ll} \cos\left( \frac{1}{x} \right) & \mbox{, if } x \neq 0 \\ 0 & \mbox{, if } x =0 \end{array} \right.$$
is discontinuous at x = 0?

They're pretty much the same. :)

------------------------------------

For short, when working with multi-variable functions.

• To prove continuity, use Squeeze Theorem.
• To prove discontinuity, choose sequence(s). :)

 

[SammyS]

...

So, to prove that a function f is discontinuous at some (x₀, y₀) you can choose 2 sequences (x_n, y_n), and (x'_n, y'_n), such that both of them tend to (x₀, y₀). But $$\lim f(x_n, y_n) \neq \lim f(x'_n, y'_n)$$.

This is fine for discontinuity and is clearer than what I suggested. The sequence {x_n} that I have in my previous post gives the sequence {f(x_n, y₀)}, where f(x_n, y₀) = (‒1)ⁿy₀, which alternates between ‒y₀ and y₀, the odd numbered members of the sequence being negative, the even being positive. Take the two sub-sequences, or take the even numbered sub-sequence (It's easier to generate.) and a sequence {f(0,y_n)}, where y_n→y₀, since f(0,y_n)=0 for all y.

Do you remember the proof that the function:
$$f:= \left\{ \begin{array}{ll} \cos\left( \frac{1}{x} \right) & \mbox{, if } x \neq 0 \\ 0 & \mbox{, if } x =0 \end{array} \right.$$
is discontinuous at x = 0?
...

Showing continuity is a bit more difficult.

The squeeze theorem is fine: Use g(x, y) = |y‒5|, for all x, and h(x, y) = ‒g(x, y).

You may also evaluate $$\textstyle \lim_{(x, y)\to(0, 5)}\,f(x,y)$$ along an arbitrary direction. It's easy to do along the line x = 0 or y = 5. Both of these obviously give zero. To approach (0, 5) from an arbitrary direction, let $$y = m\,x+5$$ and evaluate $$\textstyle \lim_{x\to 0}\,f(x,m\,x+5)$$.

This is somewhat like the function:
$$G:= \left\{ \begin{array}{ll} x\cos\left( \frac{1}{x} \right) & \mbox{, if } x \neq 0 \\ 0 & \mbox{, if } x =0 \end{array} \right.$$
which is continuous at x = 0.
Again from WolframAlphahttp://usera.ImageCave.com/DGSamSnyder/x_cos_rec_x.gif.jpg

 

[quietrain]

Do you remember the proof that the function:
$$f:= \left\{ \begin{array}{ll} \cos\left( \frac{1}{x} \right) & \mbox{, if } x \neq 0 \\ 0 & \mbox{, if } x =0 \end{array} \right.$$
is discontinuous at x = 0?

They're pretty much the same. :)

------------------------------------

For short, when working with multi-variable functions.

• To prove continuity, use Squeeze Theorem.
• To prove discontinuity, choose sequence(s). :)

the limit of cos(1/x) , x-->0, is = cos(huge number) is not equal to cos(1/0) = undefined
so it is discontinuous at x=0 right?

similarly, for (y-5)cos(1/x²)
the limit x,y --> 0,y is the same as above, so it is discontinous at x=0 and y = any value except 5

because at 5,
limit x,y --> 0,5 = 0 which is = f(0,5) = 0
so it is continuous at 0,5

so my question is , is the only way to spot 0,5 through inspection or is there a fix method of finding such "special" points?

also, since y-5 and cos(1/x² are each continuous everywhere except when x = 0,
so i conclude that the function is continuous everywhere except at (x=0 with y not equal to 5)

 

[quietrain]

The squeeze theorem is fine: Use g(x, y) = |y‒5|, for all x, and h(x, y) = ‒g(x, y).

You may also evaluate $$\textstyle \lim_{(x, y)\to(0, 5)}\,f(x,y)$$ along an arbitrary direction. It's easy to do along the line x = 0 or y = 5. Both of these obviously give zero. To approach (0, 5) from an arbitrary direction, let $$y = m\,x+5$$ and evaluate $$\textstyle \lim_{x\to 0}\,f(x,m\,x+5)$$.

the squeeze theorem is to show limit exist for a function as it approaches a point (a,b) right?
but the squeeze theorem assumes you have a conjecture about the limit itself already first before you can use it...
what if you want to FIND the limit, so that you can use the continuity theorem.

since the squeeze theorem say |f(x,y) - Limit| <or= g(x,y) , if g(x,y)tends to 0 as x,y tends to a,b...
then how do i FIND this limit first? trial and error?


like if i see the function (y-5)cos(1/x² ,
sois this the correct way to use squeeze theorem?

|(y-5)(cos1/x2)| <or= |y-5|
so y-5 is the upper bound limit
so as function tends to (0,5)
|y-5| tends to 0
so limit of f(x,y), --> 0,5 = 0
so limit f(x,y),--> (0,5) = f(0,5) = 0
so i conclude 0,5 is continous?
so i conclude that it is continuous at 0,5

 

[VietDao29]

the lim[]it of cos(1/x) , x-->0, is = cos(huge number) is not equal to cos(1/0) = undefined

f(0) = 0 as defined in f.

$$f:= \left\{ \begin{array}{ll} \cos\left( \frac{1}{x} \right) & \mbox{, if } x \neq 0 \\ 0 & \mbox{, if } x =0 \end{array} \right.$$

so it is discontinuous at x=0 right?

Yes, that's true but that's not how you prove it. You have to express your idea using Mathematical terms, like this:

Since f(0) is defined to be 0. And we know that the function is not continuous at 0. This is because cos(x) has the period of $$2 \pi$$. However, when $$x \rightarrow 0$$, $$\frac{1}{x} \rightarrow \infty$$, which means that as x tends to 0, $$\cos \left( \frac{1}{x} \right)$$ oscillates between -1, and 1. The closer x gets to 0, the faster $$\frac{1}{x}$$ grows without bounds, this leads to the fact that $$\cos \left( \frac{1}{x} \right)$$ oscillates faster.

This is how you 'imagine'. But you have to write your idea down Mathematically. Like this:

• Choose a sequence x_n that tends to 0.
• But f(x_n) does not tend to f(0) = 0 (as defined)

We choose the sequence $$x_n = \frac{1}{2n \pi}$$. Of course this sequence tends to 0, as n grows without bound, right? But we have:

$$\begin{align} f(x_n) &= \cos\left( \frac{1}{\x_n} \right) \mbox{ since } x_n \neq 0, \forall n \\ &= \cos(2n \pi)\\ &= 1 \rightarrow 1 \neq f(0) = 0 \end{align}$$

similarly, for (y-5)cos(1/x²)
the limit x,y --> 0,y is the same as above, so it is discontinous at x=0 and y = any value except 5

Yup, but can you present this Mathematically?

because at 5,
limit x,y --> 0,5 = 0 which is = f(0,5) = 0
so it is continuous at 0,5

Well, yes. But can you prove that $$\lim_{(x, y) \rightarrow (0; 5)} f(x; y) = 0$$?

so my question is , is the only way to spot 0,5 through inspection or is there a fix method of finding such "special" points?

Yes, the more you practice, the better you'll get at spotting out these special points. One method to spot out these point has been mentioned earlier in the post. Read it carefully, and see if you can get the point.

You may also evaluate $$\textstyle \lim_{(x, y)\to(0, 5)}\,f(x,y)$$ along an arbitrary direction. It's easy to do along the line x = 0 or y = 5. Both of these obviously give zero. To approach (0, 5) from an arbitrary direction, let $$y = m\,x+5$$ and evaluate $$\textstyle \lim_{x\to 0}\,f(x,m\,x+5)$$.

No, this is not arbitrary enough. (x; y) not only tends to (0; 5) in straight lines, it can also tends to (0; 5) in many curves.

You can consider this example:
$$f(x; y) := \left\{ \begin{array}{ll} \frac{x ^ 3 y}{x ^ 6 + y ^ 2} & \mbox{ , if } (x; y) \neq (0; 0) \\ 0 & \mbox { , if } (x; y) = (0; 0) \end{array} \right.$$

If (x; y) tends to (0; 0) along straight lines, i.e y = mx.
We have:
$$f(x, mx) = \frac{mx ^ 4}{x ^ 6 + m ^ 2 x ^ 2} = \frac{mx ^ 2}{x ^ 4 + m ^ 2} \rightarrow 0$$
However, if we go along the curve y = x³, we have:
$$f(x, x ^ 3) = \frac{x ^ 6}{x ^ 6 + x ^ 6} = \frac{1}{2} \rightarrow \frac{1}{2}$$.

 

[SammyS]

You may also evaluate $$\textstyle \lim_{(x, y)\to(0, 5)}\,f(x,y)$$ along an arbitrary direction. It's easy to do along the line x = 0 or y = 5. Both of these obviously give zero. To approach (0, 5) from an arbitrary direction, let $$y = m\,x+5$$ and evaluate $$\textstyle \lim_{x\to 0}\,f(x,m\,x+5)$$.

No, this is not arbitrary enough. (x; y) not only tends to (0; 5) in straight lines, it can also tends to (0; 5) in many curves.

You can consider this example:
$$f(x; y) := \left\{ \begin{array}{ll} \frac{x ^ 3 y}{x ^ 6 + y ^ 2} & \mbox{ , if } (x; y) \neq (0; 0) \\ 0 & \mbox { , if } (x; y) = (0; 0) \end{array} \right.$$

If (x; y) tends to (0; 0) along straight lines, i.e y = mx.
We have:
$$f(x, mx) = \frac{mx ^ 4}{x ^ 6 + m ^ 2 x ^ 2} = \frac{mx ^ 2}{x ^ 4 + m ^ 2} \rightarrow 0$$
However, if we go along the curve y = x³, we have:
$$f(x, x ^ 3) = \frac{x ^ 6}{x ^ 6 + x ^ 6} = \frac{1}{2} \rightarrow \frac{1}{2}$$.

Of course VietDao29 is correct. I took a short-cut that was erroneous.

Very nice counter-example to show my mistake!

 

[quietrain]

f(0) = 0 as defined in f.

$$f:= \left\{ \begin{array}{ll} \cos\left( \frac{1}{x} \right) & \mbox{, if } x \neq 0 \\ 0 & \mbox{, if } x =0 \end{array} \right.$$

oh so i am suppose to check my limits with what the question specifies? like if question specifies when x =0 , f(x,y) = cos(infinity)
then when i let x -> 0, it is indeed cos(infinity) so it is continuous in this case?

ok about the proving part

i just need to follow your guidelines

but i choose x = sqrt( 1/ 2nPI) ----- (why is there a n here? for generalization to all values of cos? i am suppose to let x tend to 0 right? but if i let n tend to infinity i effectively let x tend to 0? so now i am playing with 'n'?

so i get (y-5)(cos(2nPI))
when (x,y) -> (0,y)
= (y-5)(1) =/= f(0,y) = 0 unless y=5

so is this ok?

 

[VietDao29]

oh so i am suppose to check my limits with what the question specifies? like if question specifies when x =0 , f(x,y) = cos(infinity)
then when i let x -> 0, it is indeed cos(infinity) so it is continuous in this case?

ok about the proving part

i just need to follow your guidelines

but i choose x = sqrt( 1/ 2nPI) ----- (why is there a n here? for generalization to all values of cos? i am suppose to let x tend to 0 right? but if i let n tend to infinity i effectively let x tend to 0? so now i am playing with 'n'?

so i get (y-5)(cos(2nPI))
when (x,y) -> (0,y)
= (y-5)(1) =/= f(0,y) = 0 unless y=5

so is this ok?

Do you known about sequences? n is just the index of the term in a sequence.

Yup, your proof looks good. What you are doing here is to show that there exist a sequence (x_n, y_n) tends to (0; y), but f(x_n, y_n) does not tend to f(0; y) = 0.

--------------------

When y = 5, you should use the Squeeze Theorem to prove that the function is continuous at that (0; 5). Let's try to see if you can do it. :)

 

[quietrain]

|(y-5)(cos1/x2)| <or= |y-5|
so y-5 is the upper bound limit
so as function tends to (0,5)
|y-5| tends to 0
so limit of f(x,y), --> 0,5 = 0
so limit f(x,y),--> (0,5) = f(0,5) = 0
so i conclude 0,5 is continous?
so i conclude that it is continuous at 0,5

like this?

about sequences, not sure... :( you mean those power series ? )

 

[VietDao29]

|(y-5)(cos1/x2)| <or= |y-5|
so y-5 is the upper bound limit
so as function tends to (0,5)
|y-5| tends to 0
so limit of f(x,y), --> 0,5 = 0
so limit f(x,y),--> (0,5) = f(0,5) = 0

Your notation looks weird. What do the last 2 lines mean?

And, about the definition of sequence, you should look again at your textbook. You learned about sequence, limit of a sequence, limit of a function a while back in High School. When working with limit (for one or multi-variable functions), you'll need to use these concepts a lot. So, a little bit of revision will probably help you. You may also Google to search for more information.

Here's 3 wiki pages that may help you:

• http://en.wikipedia.org/wiki/Sequence_%28mathematics%29" [Broken]
• http://en.wikipedia.org/wiki/Limit_of_a_sequence" [Broken]
• http://en.wikipedia.org/wiki/Limit_of_a_function" [Broken]

 

[quietrain]

the last 2 lines mean

limit of f(x,y) as (x,y) tends to (0,5) = 0 (found limit of f through squeeze theorem here)

so by continuity theorem

f(0,5) = 0

so since the limit of f as it tends to 0,5 is 0, and f(0,5) = 0, hence i conclude function is continuous at 0,5 ?

i will have a look at those 3 long links :( thanks anyway!

 

[VietDao29]

the last 2 lines mean

limit of f(x,y) as (x,y) tends to (0,5) = 0 (found limit of f through squeeze theorem here)

so by continuity theorem

f(0,5) = 0

so since the limit of f as it tends to 0,5 is 0, and f(0,5) = 0, hence i conclude function is continuous at 0,5 ?

i will have a look at those 3 long links :( thanks anyway!

Looks all good now. Congratulations. :)

 

[quietrain]

thanks!