• Support PF! Buy your school textbooks, materials and every day products Here!

Did I do this correctly? version: limits in 3 variables

  • Thread starter ggcheck
  • Start date
  • #1
87
0

Homework Statement

evaluate the limit or determine that it does not exist.

lim [sin(x)/sin(y)]
(x,y)--->(pi,0)

The Attempt at a Solution


since it is not continues at point (pi,0) I can't use use substitution, so I attempted to prove that the limit does not exist by evaluating it along the x and y axises

along x-axis:
lim [sin(x)/sin(0)] = DNE
(x,y)-->(x,0)

along y-axis:
lim [sin(0)/sin(y)] = 0
(x,y)-->(0,y)

since the two are not equal, the limit does not exist

is this correct? am I even close :(
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,793
922
The only question I would have is "why did you put x= 0?". The limit is at x= [itex]\pi[/itex].

But you are correct that since the function is undefined at any point on the x-axis, the limit at ([itex]\pi[/itex], 0) cannot exist.
 
  • #3
87
0
I'm not really sure what I'm doing... we didn't cover this in class, and the book doesn't have any other similar examples
 
  • #4
87
0
any tips/ways to approach similar problems?
 
  • #5
Defennder
Homework Helper
2,591
5
Actually your method is correct, as Halls pointed out. The only problem is that in here:

along x-axis:
lim [sin(x)/sin(0)] = DNE
(x,y)-->(x,0)
Replace with y=0 [tex]lim_{x \rightarrow \pi}[/tex]

along y-axis:
lim [sin(0)/sin(y)] = 0
(x,y)-->(0,y)
Replace with "along line [tex]x=\pi[/tex] and [tex]lim_{y \rightarrow 0}[/tex]"
 
  • #6
87
0
so I'd prove that the original limit doesn't exist by showing:

lim[sin(x)/sin(y)] = [sin(pi)/sin(y)] = 0
(x,y)-->(pi,y)

lim[sin(x)/sin(y)] = [sin(pi)/sin(0)] DNE
(x,y)-->(pi,0)
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,793
922
Actually the second of those is sufficient: if the limit along a single path "does not exist" then the limit itself cannot exist.

More generally, however, a limit exists if and only if the limit along all possible paths exists and is the same for all possible paths. If you can show that two different paths give different limits, then the limit itself does not exist.

Of course, you can't use that to prove that a limit does exist, since you can't check all possible paths. In that case, with (0,0) as the limit point, I would recommend changing to polar coordinates where the distance from (0,0) is measured by the single variable r. If the limit as r goes to 0 is independent of [itex]\theta[/itex], then the limit exists.

Which brings up my final question: why did you title this "limits in 3 variables" when your problem has only 2 variables?
 
  • #8
87
0
"Which brings up my final question: why did you title this "limits in 3 variables" when your problem has only 2 variables?"
I misspoke

How about this: Since I have a hunch the limit doesn't exist I prove that it doesn't exist by showing that you get two different limits when evaluating over separate paths

the original:

lim[sin(x)/sin(y)] = [sin(pi)/sin(y)] = 0
(x,y)-->(pi,y)

2nd one, with new paths:
y = x - pi/2
x = y + pi/2

lim[sin(y+{pi/2})/sin(x-{pi/2})]= [sin(pi/2)/sin(pi/2)] = 1/1
(x,y)-->(pi,0)

yes/no?
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,793
922
If x goes to pi and y= x- pi/2, then y goes to pi/2, not 0.
 
  • #10
87
0
I think I figured out the 2nd part:

x= y + pi

lim[sin(y + pi)/sin(y)] = [sin(y)cos(pi)+cos(y)sin(pi)]/sin(y) = cos(pi) + cot(y)sin(pi) = 1 +0
(x,y)-->(pi,0)
 
  • #11
HallsofIvy
Science Advisor
Homework Helper
41,793
922
cos(pi) is not equal to 1!
 

Related Threads for: Did I do this correctly? version: limits in 3 variables

  • Last Post
Replies
1
Views
1K
Replies
3
Views
2K
Replies
5
Views
644
  • Last Post
Replies
2
Views
810
Replies
9
Views
1K
Replies
2
Views
424
Replies
3
Views
18K
Replies
1
Views
1K
Top