OpenStudy (dbwong):

limits of two variables question

1 year ago
OpenStudy (dbwong):

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

1 year ago
OpenStudy (johnweldon1993):

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.

1 year ago
OpenStudy (johnweldon1993):

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!

1 year ago
OpenStudy (johnweldon1993):

This was working off of the work done here: http://www.math.tamu.edu/~glahodny/Math251/Section%2012.2.pdf

1 year ago
OpenStudy (holsteremission):

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.

1 year ago