Limit & Domain Homework: Find Solution

In summary: I am not sure how to change x-1 into polar coordinates. Is there a hint you can give me to point me in the right direction?Thanks for your replies, I am a bit rusty with polar coordinates but using y=mx I get the following with the function. \frac {(x-1)xm^2x^2}{x^2+m^2x^2} \frac {(x-1)xm^2x^2}{x^2(m^2+1)} \frac {(x-1)xm^2}{m^2+1} \frac {m^2x^2-m^2x}{m^2+1}is this the right method? also having
  • #1
bayan
203
0

Homework Statement



Find the limit (if it exists) and domain of the following function. (I do not need to use epsilon delta proof, just results on different curves to prove this)

[itex] \stackrel{lim}{_{(x,y)→(−1,0)}}[/itex] [itex] \frac {x(x+1)y^2}{(x+1)^2+y^2}[/itex]

Homework Equations



The Attempt at a Solution

Just by looking at function I can see that Domain is [itex]ℝ | x≠0, x≠-1, y≠0 , y^2≠-(x+1)^2|[/itex]

Having graphed that function in Google I can see that at [itex](-1,0)[/itex] the function is not continuous and if I try and substitute the x and y values into function I end up with [itex]\frac{0}{0}[/itex] and I get similar results if I try [itex]y=mx[/itex]

I have also tried
[itex] \stackrel{lim}{_{x→0}}[/itex] [itex] \frac {(x-1)(x-1+1)y^2}{(x-1+1)^2+y^2}[/itex] and [itex]y=mx[/itex]

Regards
 
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  • #2
You may want to look closer at the function to find the domain, for example why is x = 0 not allowed, or y² = (x + 1)² ?

If you substitute y = mx and let x go to -1 you should not get 0/0. Can you show us your calculation in a little more detail?
 
  • #3
bayan said:

Homework Statement



Find the limit (if it exists) and domain of the following function. (I do not need to use epsilon delta proof, just results on different curves to prove this)

[itex] \stackrel{lim}{_{(x,y)→(−1,0)}}[/itex] [itex] \frac {x(x+1)y^2}{(x+1)^2+y^2}[/itex]

Homework Equations



The Attempt at a Solution



Just by looking at function I can see that Domain is [itex]ℝ | x≠0, x≠-1, y≠0 , y^2≠(x+1)^2|[/itex]

Having graphed that function in Google I can see that at [itex](-1,0)[/itex] the function is not continuous and if I try and substitute the x and y values into function I end up with [itex]\frac{0}{0}[/itex] and I get similar results if I try [itex]y=mx[/itex]

I have also tried
[itex] \stackrel{lim}{_{x→0}}[/itex] [itex] \frac {(x-1)(x-1+1)y^2}{(x-1+1)^2+y^2}[/itex] and [itex]y=mx[/itex]

Regards
x=0, does not make the denominator zero. Neither does y2 = (x+1)2 make the denominator zero.

In fact when x = -1, the only case for which the denominator is zero, is when you also have y = 0

Of course, that's precisely the point at which you need to find the limit.

By the way, the line given by y = mx does not pass through the point, (-1, 0) . Consider the line y = m(x+1) . Looking at the limit along this line is only helpful if the limit doesn't exist.


I suggest shifting you curve one unit to the right., and the switching to polar coordinates.
 
  • #4
Thanks for your replies,

I missed a - sign in my OP, so the denominator goes to zero if [itex]x=-1[/itex] [itex]y=0[/itex] or [itex]y^2=-(x+1)^2[/itex] and numerator goes to zero if [itex]x=0[/itex] or [itex]x=-1[/itex]


By shifting curve to the right I get the following for the function [itex]\frac{(x-1)xy^2}{x^2+y^2}[/itex] and by letting [itex]y=m(x+1)[/itex] I get [itex] \frac {(x-1)xm^2(x+1)^2}{x^2+m^2(x+1)^2} [/itex] I can't seem to take it any further from here on and subbing 0 for [itex] x[/itex] I get [itex] \frac {0}{m^2}[/itex]

where have I gone wrong in try to find the limit ( or showing that it does not exist)?

Regards
 
Last edited:
  • #5
For [itex]m \neq 0[/itex], 0/m² = 0.
All you need to check is that it is also 0 when y = 0, i.e. where you are approaching along the x-axis.

I thought you were talking about the function over the reals by the way, in which case [itex]y^2 = -(x + 1)^2[/itex] will never hold (except for when x = -1 and y = 0 as you already noted), because y² and (x + 1)² are always positive.
 
  • #6
bayan said:
Thanks for your replies,

I missed a - sign in my OP, so the denominator goes to zero if [itex]x=-1[/itex] [itex]y=0[/itex] or [itex]y^2=-(x+1)^2[/itex] and numerator goes to zero if [itex]x=0[/itex] or [itex]x=-1[/itex]
Which sign did you change?

By shifting curve to the right I get the following for the function [itex]\frac{(x-1)xy^2}{x^2+y^2}[/itex] and by letting [itex]y=m(x+1)[/itex] I get [itex] \frac {(x-1)xm^2(x+1)^2}{x^2+m^2(x+1)^2} [/itex] I can't seem to take it any further from here on and subbing 0 for [itex] x[/itex] I get [itex] \frac {0}{m^2}[/itex]

where have I gone wrong in try to find the limit ( or showing that it does not exist)?

Regards
Now that you've shifted the function you need to do the limit of the shifted function as (x, y) (0, 0), so you should use ##\ y = mx\ .\ ## Unless I've messed something up, the resulting limit does exist. That being the case, the problem may appear to be solved, but it's not solved. Such a method can only show (in some cases) that a limit does not exist.

Change ##\displaystyle \ \frac{(x-1)xy^2}{x^2+y^2}\ ## to polar coordinates, for which you merely need to consider the limit as r → 0 .
 
  • #7
SammyS said:
Which sign did you change?


Now that you've shifted the function you need to do the limit of the shifted function as (x, y) (0, 0), so you should use ##\ y = mx\ .\ ## Unless I've messed something up, the resulting limit does exist. That being the case, the problem may appear to be solved, but it's not solved. Such a method can only show (in some cases) that a limit does not exist.

Change ##\displaystyle \ \frac{(x-1)xy^2}{x^2+y^2}\ ## to polar coordinates, for which you merely need to consider the limit as r → 0 .

Thanks for your replies, I am a bit rusty with polar coordinates but using [itex]y=mx[/itex] I get the following with the function.

[itex] \frac {(x-1)xm^2x^2}{x^2+m^2x^2}[/itex]

[itex] \frac {(x-1)xm^2x^2}{x^2(m^2+1)} [/itex]

[itex] \frac {(x-1)xm^2}{m^2+1}[/itex]

[itex] \frac {m^2x^2-m^2x}{m^2+1}[/itex]


is this the right method? also having a quick read on polar coordinates I see that [itex]x^2+y^2=r^2[/itex] how can I change [itex] (x-1)xy^2[/itex] to polar form?

Regards
 
  • #8
The standard conversion to polar coordinates is:
x = r cos(θ)
y = r sin(θ)
from which we get x2 + y2 = r2

On the line y = mx, y = m*r*cos(θ).

BTW, instead of using SIZE to make itex stuff larger, just use tex without the SIZE tags. It's also better, in my opinion, to use the least amount of LaTeX that you can, as large tracts of LaTeX can bog some browsers down, making the pages slow to render. None of what I wrote above uses LaTeX.
 
  • #9
bayan said:
Thanks for your replies, I am a bit rusty with polar coordinates but using [itex]y=mx[/itex] I get the following with the function.

[itex] \frac {(x-1)xm^2x^2}{x^2+m^2x^2}[/itex]

[itex] \frac {(x-1)xm^2x^2}{x^2(m^2+1)} [/itex]

[itex] \frac {(x-1)xm^2}{m^2+1}[/itex]

[itex] \frac {m^2x^2-m^2x}{m^2+1}[/itex]


is this the right method? also having a quick read on polar coordinates I see that [itex]x^2+y^2=r^2[/itex] how can I change [itex] (x-1)xy^2[/itex] to polar form?

Regards
Have you taken the limit of [itex]\displaystyle \ \frac {m^2x^2-m^2x}{m^2+1}\ [/itex] as x → 0 ?

While approaching (0, 0) along the line y = mx can be useful to show that a limit doesn't exist, it can't show that a limit (x,y) → (0,0) does exist, although if the limit does exist, this will show you what the limit is.

When you change to polar coordinates do it with the expression ## \displaystyle \ \frac{(x-1)xy^2}{x^2+y^2}\,,\ ## not with [itex]\displaystyle \ \frac {m^2x^2-m^2x}{m^2+1}\ .[/itex]
 
  • #10
SammyS said:
Have you taken the limit of [itex]\displaystyle \ \frac {m^2x^2-m^2x}{m^2+1}\ [/itex] as x → 0 ?

While approaching (0, 0) along the line y = mx can be useful to show that a limit doesn't exist, it can't show that a limit (x,y) → (0,0) does exist, although if the limit does exist, this will show you what the limit is.

When you change to polar coordinates do it with the expression ## \displaystyle \ \frac{(x-1)xy^2}{x^2+y^2}\,,\ ## not with [itex]\displaystyle \ \frac {m^2x^2-m^2x}{m^2+1}\ .[/itex]

Thanks, so in polar coordinates the function can be written as [itex]\displaystyle \ \frac{(rcosθ-1)rcosθ(rsinθ)^2}{r^2}[/itex]

and just subbing [itex]\displaystyle r=0[/itex] yields to [itex]\displaystyle \frac {0}{0}[/itex] or do I need to further simplify the function in polar form and then sub in [itex]\displaystyle r=0[/itex]

Also for the domain I can see as you mentioned before (for the denominator part) [itex]\displaystyle (x+1)^2[/itex] can not be a negative number as it is square and same applies to [itex]\displaystyle y^2[/itex] would the domain be all [itex]\displaystyle ℝ[/itex] , [itex]\displaystyle y≠0[/itex] or [itex]\displaystyle x≠-1,0[/itex]

Regards
 
  • #11
bayan said:
Thanks, so in polar coordinates the function can be written as [itex]\displaystyle \ \frac{(rcosθ-1)rcosθ(rsinθ)^2}{r^2}[/itex]

and just subbing [itex]\displaystyle r=0[/itex] yields to [itex]\displaystyle \frac {0}{0}[/itex] or do I need to further simplify the function in polar form and then sub in [itex]\displaystyle r=0[/itex]

Also for the domain I can see as you mentioned before (for the denominator part) [itex]\displaystyle (x+1)^2[/itex] can not be a negative number as it is square and same applies to [itex]\displaystyle y^2[/itex] would the domain be all [itex]\displaystyle ℝ[/itex] , [itex]\displaystyle y≠0[/itex] or [itex]\displaystyle x≠-1,0[/itex]

Regards
How much work have you done with limits?

What sort of answer is 0/0 for a limit or for anything else?

Factor whatever power of r that you can from the numerator & cancel whatever possible from the denominator.

How much work have you done with limits?
 
  • #12
SammyS said:
How much work have you done with limits?

a little bit, but this one is confusing me for some reason (possibly lack of experience with limits)


SammyS said:
What sort of answer is 0/0 for a limit or for anything else?

## \frac {0}{0} ## would be undermined, right?

SammyS said:
Factor whatever power of r that you can from the numerator & cancel whatever possible from the denominator.

so this is how I have gone with factoring it:

## \frac{(r cos θ -1)r cos θ (r sin θ)^2}{r^2}
= \frac {(r^2 cos^2 θ - r cos θ)r^2 sin ^2 θ}{r^2}
= \frac {r((r cos^2 θ - cos θ)r sin^2 θ)}{r^2}
= \frac {(r cos^2 θ - cos θ)r sin^2 θ}{r}
= (r cos^2 θ - cos θ) sin^2 θ ##

and setting ##r=0## i get ## -cos θ sin^2 θ## do I then use trig identity ## sin^2 θ = \frac {1-cos 2θ}{2}## ? also as ##r→0## does ##θ## also tend to ##0##?

Regards
 
  • #13
##\displaystyle \ \frac{(rcosθ-1)rcosθ(rsinθ)^2}{r^2}##

is the same as

##\displaystyle \ \frac{r^3(r\cosθ-1)\cosθ(\sin^2θ)}{r^2}\quad\to\quad r(r\cosθ-1)\cosθ( \sin^2θ)\ .##

As r → 0, this is 0, for all values of θ . Right ?
 
  • #14
SammyS said:
As r → 0, this is 0, for all values of θ . Right ?


Thanks, so the limit is at [itex](-1,0) [/itex] is 0 regardless of the path taken ( which would be represented by θ) right?

also did I have the right idea in regards to the domain of the funtion? all ##ℝ## ##|y≠0## , ##x≠-1,0|##
Regards
 
  • #15
bayan said:
Thanks, so the limit is at [itex](-1,0) [/itex] is 0 regardless of the path taken ( which would be represented by θ) right?

also did I have the right idea in regards to the domain of the funtion? all ##ℝ## ##|y≠0## , ##x≠-1,0|##
Regards
The original function, ##\displaystyle \ \frac {x(x+1)y^2}{(x+1)^2+y^2}\ ##, is defined at x=0 and y=0, i.e. at (x,y)=(0,0), and is 0 at that point.

##\displaystyle \frac{0(0+1)0^2}{(0+1)^2+0^2}=\frac{0}{1}=0\ ##
 
  • #16
Given that (0,0) is defined would I be correct in saying the function in undefined only at (-1,0) hence ##D= R## except ##x=-1, y=0## ?

Regards
 
  • #17
bayan said:
Given that (0,0) is defined would I be correct in saying the function in undefined only at (-1,0) hence ##D= R## except ##x=-1, y=0## ?

Regards

\mathbb{R}

The domain is all of ##\ \mathbb{R}^2\,,\ ## except (-1, 0) .
 
  • #18
Thanks for your help
 

1. What is the purpose of the "Limit & Domain Homework: Find Solution" assignment?

The purpose of this assignment is to help students understand the concept of limits and domains in mathematics. It also allows them to practice finding solutions to problems involving limits and domains.

2. What is a limit in mathematics?

A limit is a value that a function or sequence approaches as the input or variable approaches a certain value. It is used to describe the behavior of a function or sequence near a specific input or value.

3. What is a domain in mathematics?

The domain of a function is the set of all possible input values for which the function is defined. It is the set of all values that the independent variable can take on in a function.

4. How do I find the limit of a function?

To find the limit of a function, you can either evaluate the function at the specific input value, or use algebraic techniques such as factoring or rationalizing to simplify the function and then evaluate the limit. You can also use a graph or table to estimate the limit.

5. What is the importance of understanding limits and domains in mathematics?

Understanding limits and domains is essential in calculus and other advanced mathematics courses. It allows us to analyze the behavior of functions and make predictions about their values. It also helps us determine the validity of mathematical expressions and solve real-world problems involving rates of change and optimization.

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