Does the limit of (sin(x)sin(x))/(x^2 + y^2) as (x,y) approaches (0,0) exist?

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In summary: To give the example:$$\lim_{t \to 0} \frac{\sin(t)\sin(0)}{t^2+0}= \lim_{t \to 0} \frac{\sin(t)\cdot 0}{t^2}= \lim_{t \to 0} \frac{0}{t^2}= \lim_{t \to 0}\ 0 = 0$$See how I postpone filling in 0 until I no longer have a result that would lead to 0/0?
  • #1
Jamie2
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I need help showing that the
limit as (x,y)-->(0,0) of (sin(x)sin(x))/x2+y2 does not exist.

I've tried approaching the function along the path y=mx, x=y, y=x^3, and several other paths and am not getting any different limits.
 
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  • #2
Jamie said:
I need help showing that the
limit as (x,y)-->(0,0) of (sin(x)sin(x))/x2+y2 does not exist.

I've tried approaching the function along the path y=mx, x=y, y=x^3, and several other paths and am not getting any different limits.

Welcome to MHB, Jamie! :)

Which limits do you get?

Easiest should be x=0 versus y=0.
 
  • #3
I like Serena said:
Welcome to MHB, Jamie! :)

Which limits do you get?

Easiest should be x=0 versus y=0.

for both x=0 and y=0 I got that the limit is undefined (0/0)
 
  • #4
Jamie said:
for both x=0 and y=0 I got that the limit is undefined (0/0)

Hmm, I am not getting (0/0)...
How did you arrive at that? :confused:

Let's pick one. With x=0 we get:
$$\lim_{y \to 0} \frac{\sin(0)\sin(0)}{0^2+y^2}
= \lim_{y \to 0} \frac{0}{y^2}
= \lim_{y \to 0} 0
= 0$$

How far can you get with y=0?
Can you perhaps show some of your workings?
 
  • #5
I like Serena said:
Hmm, I am not getting (0/0)...
How did you arrive at that? :confused:

Let's pick one. With x=0 we get:
$$\lim_{y \to 0} \frac{\sin(0)\sin(0)}{0^2+y^2}
= \lim_{y \to 0} \frac{0}{y^2}
= \lim_{y \to 0} 0
= 0$$

How far can you get with y=0?
Can you perhaps show some of your workings?

sorry! i made a typo. the original problem has sin(x)sin(y) in the numerator
 
  • #6
Jamie said:
sorry! i made a typo. the original problem has sin(x)sin(y) in the numerator

Okay... what do you get for x=0 then?
And can you show some of your thoughts for y=x?
 
  • #7
I like Serena said:
Okay... what do you get for x=0 then?
And can you show some of your thoughts for y=x?

along the path y=x
x=t
y=t

limit as t-->0 of sint*sint/t^2+t^2 = sin(0)sin(0)/(0^2+0^2)= 0/0=undefined

along the path x=0
x=0
y=t

limit as t-->0 sin(t)*sin(0)/t^2+0= sin(0)*sin(0)/0^2+0=0/0 =undefined
 
  • #8
Jamie said:
along the path x=0
x=0
y=t

limit as t-->0 sin(t)*sin(0)/t^2+0= sin(0)*sin(0)/0^2+0=0/0 =undefined

Let's start with this one.

When you take a limit, the variable approaching 0 ($t$ in this case) is considered to be non-zero at all times. That is part of the definition of a limit. When you would find 0/0 when you would fill in 0, that means that you are not done yet.

To give the example:
$$\lim_{t \to 0} \frac{\sin(t)\sin(0)}{t^2+0}
= \lim_{t \to 0} \frac{\sin(t)\cdot 0}{t^2}
= \lim_{t \to 0} \frac{0}{t^2}
= \lim_{t \to 0}\ 0 = 0$$
See how I postpone filling in 0 until I no longer have a result that would lead to 0/0?
 

FAQ: Does the limit of (sin(x)sin(x))/(x^2 + y^2) as (x,y) approaches (0,0) exist?

What is the definition of a limit?

The limit of a function, f(x), at a point, a, is the value that f(x) approaches as x gets closer and closer to a. In mathematical notation, this can be written as lim x→a f(x).

How can you prove that a limit does not exist?

A limit does not exist if the left-hand limit and the right-hand limit do not have the same value, or if either the left-hand limit or the right-hand limit is undefined.

What does it mean if a limit is undefined?

If a limit is undefined, it means that the function approaches different values from the left and right sides of the point. This can happen if there is a jump or a discontinuity at that point, or if the function is undefined at that point.

Can a limit not exist if the function is continuous?

Yes, a limit can still not exist even if the function is continuous. This can happen if the function is undefined at that point, or if there is a jump or a discontinuity in the function at that point.

Are there any other ways to prove that a limit does not exist?

Yes, there are other ways to prove that a limit does not exist, such as using the epsilon-delta definition of a limit or using the Squeeze Theorem. It ultimately depends on the specific function and the situation at hand.

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