Question

limits of two variables question

Answer 1

for number 6, i just dont see what method to use... help? tip to start?

Answer 2

Hmm, never actually done these before

But basically, remember a limit definition of just 1 variable
\[\large \lim_{x\rightarrow a} f(x)\]

This limit exists if, and only if, \[\large \lim_{x\rightarrow a^{-}} f(x) = \lim_{x\rightarrow a^{+}} f(x) \]

That was easy, since we are just approaching a point along a path from both directions

In THIS case...we have a function of 2 variables...and how many paths can we follow to get to that point? An infinite number of paths will get you to your point

So there is a theorem that states:
If \(\large f(x,y) \rightarrow L_1\) along a path \(\large P_1 \) and \(\large f(x,y) \rightarrow L_2\) along a path \(\large P_2 \) as \(\large (x,y) \rightarrow (a,b) \) where \(\large L_1 \cancel{=} L_2 \) Then the limit does not exist.

Answer 3

So for your question

\[\large \lim_{(x,y) \rightarrow (0,0)} \frac{xycos(y)}{8x^2 + y^2}\]

We pick a RANDOM path (Function) for 'x' or 'y' and use that to simplify this down

So lets say I choose the path \(\large x = y\)

If I now plug in 'y' for every 'x' I see...we have

\[\large \lim_{(x,y) \rightarrow (0,0)} \frac{y^2cos(y)}{8y^2 + y^2}\]
Now simplify it down
\[\large \lim_{(y) \rightarrow (0)} \frac{\cancel{y^2}cos(y)}{\cancel{9y^2}}\]
Giving you the result of \(\large 1\)

Now if we start over and choose any random path for 'y'...we see if we arrive back at 1 for the result...otherwise the limit will not exist

Lets NOW say \(\large y = 1\)

\[\large \lim_{(x,y) \rightarrow (0,0)} \frac{xycos(y)}{8x^2 + y^2}\]
Plugging that in:
\[\large \lim_{(x) \rightarrow (0)} \frac{xcos(1)}{8x^2 + 1}\]
Now as x goes to 0 we arrive at the result \(\large 0\)

Meaning, since we got 2 different limits...The limit does not exist!

Answer 4

This was working off of the work done here:

http://www.math.tamu.edu/~glahodny/Math251/Section%2012.2.pdf

Answer 5

We can't choose \(y=1\) as a possible path because that line doesn't pass through the origin (\(y\) must be approaching \(0\) in some way).

To ensure that \((x,y)\to(0,0)\), a better choice of path might be any line through the origin \(y=mx\), or any power function of the form \(y=x^m\). If \(y=mx\), then
\[\lim_{(x,y)\to(0,0)}\frac{xy\cos y}{8x^2+y^2}=\lim_{x\to0}\frac{mx^2\cos(mx)}{8x^2+(mx)^2}=\lim_{x\to0}\frac{m\cos(mx)}{8+m^2}=\frac{m}{8+m^2}\]which tells us the limit depends on \(m\), so it can't exist.