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Show that limit does not exist

Jamie

New member
Feb 11, 2014
17
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
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.
 

Jamie

New member
Feb 11, 2014
17
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
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?
 

Jamie

New member
Feb 11, 2014
17
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
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
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?
 

Jamie

New member
Feb 11, 2014
17
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
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,908
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?