My Math Forum Limits of Functions of Two Variables

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 February 24th, 2017, 04:08 PM #1 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Limits of Functions of Two Variables In Calculus III, we've covered limits of functions of two variables, but I'm still somewhat confused by the topic. The limit has been defined to only exist if the limit approaches the same value from "all possible paths" along the surface. However, we never definitively showed how to test "all possible paths." We could show that a limit might not exist at a particular point by evaluating the limit along different paths and observing that the values of those limits are different, but showing that a limit doesn't exist at a point is much easier than showing that it does. So for any given function of two variables, what is the process to show that a limit exists for sure at a point on the surface (or a point where the function is undefined)?
 February 24th, 2017, 06:45 PM #2 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 If the function is continuous at the point, you can just plug in the values. Otherwise you could convert to polars. If you want a more rigorous proof you can use the precise definition: If for all \displaystyle \begin{align*} \epsilon > 0 \end{align*} there exists a \displaystyle \begin{align*} \delta > 0 \end{align*} such that \displaystyle \begin{align*} \sqrt{ \left( x - a \right) ^2 + \left( y - b \right) ^2 } < \delta \implies \left| f\left( x, y \right) - L \right| < \epsilon \end{align*} then \displaystyle \begin{align*} \lim_{\left( x, y \right) \to \left( a , b \right) } f\left( x,y \right) = L \end{align*}.
 February 24th, 2017, 06:51 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond In other words, there is no distinct method that can be applied to any function to determine whether or not a limit exists. Thanks from v8archie
 February 24th, 2017, 07:25 PM #4 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Okay, so let's take this, for example: So if we're to show that this piecewise function is continuous, we have to show that the limit as (a,b) approaches (0,0) of the top portion of the piecewise function is zero from all directions. I'm still not sure though; how exactly would I do that?
February 24th, 2017, 08:45 PM   #5
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Quote:
 Originally Posted by John Travolski Okay, so let's take this, for example: So if we're to show that this piecewise function is continuous, we have to show that the limit as (a,b) approaches (0,0) of the top portion of the piecewise function is zero from all directions. I'm still not sure though; how exactly would I do that?
convert to polars

$\begin{cases}\dfrac{r^2 \sin(2\theta)}{2 r^2} &r \neq 0 \\0 &r=0\end{cases} =$

$\begin{cases}\dfrac{ \sin(2\theta)}{2} &r \neq 0 \\0 &r=0 \end{cases}$

This will have a continuum of values as $0 \leq \theta < 2\pi$ and thus there will be no limit at $(0,0)$

 February 25th, 2017, 09:31 AM #6 Senior Member   Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 Okay, I think that I understand now. Thank you.
 February 25th, 2017, 02:11 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra More easily, approach the origin along the line $y=kx$ and we clearly approach $\frac{k}{1+k^2}$ which is equal to zero only when $k=0$. Thanks from agentredlum and John Travolski Last edited by v8archie; February 25th, 2017 at 02:22 PM.

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