Graphing a two variable limit

Hi, I am new to MATLAB and I can't seem to figure out how to graph a two variable limit like:

I've seen answers on how to create a two-variable function, or how check the limit for continuity, but I hope to see it graphed in relation to . I have gotten errors saying that the function is too complex, or that Z must be a matrix, and I have no clue how I should handle those kinds of errors.

Thanks in advance for any help that is given!

[Torsten]

What do you want to graph here ? The function is defined and continuous at (0,1) with value (acos(0))/1 = pi/2.

Answer 1

[Chunru]

x= -0.5:.01:0.5;
y = 0.5:.01:1.5;
[xx, yy] = meshgrid(x, y);
zz = acos(xx./yy)./(1+xx.*yy);
%zz = nan(size(xx));
% idx = abs(xx./yy)<=1;
% zz(idx) = acos(xx(idx)./yy(idx))./(1+xx(idx).*yy(idx));
% whos
contourf(xx, yy, zz);
hold on
plot(0, 1, 'rd')

[Stephen]

Thanks, that answer works, but I should of mentioned that I hoped to graph the function using something like mesh(), sorry about that.

I've tried:

x=-3:0.1:3;
y=-3:0.1:3;
[x,y]=meshgrid(x,y)
z = (acos(x./y))/((1+x.*y));
mesh(x,y,z)

but due to it being a limit it returns:

from the discontinuity.

Is it possible at all to represent that using a mesh (or any 3D graphing functions for that matter) or no due to the discontinuity?

Thank you again for the help!

[Torsten]

There is no problem with the point (0,1). It's a point of continuity - the function is even analytic there.

The only thing to remember in a graphical representation is that x must be smaller than y since acos(z) is real-valued only for abs(z) <= 1.

[Stephen]

I understand that, I was just asking the wrong question originally and got the correct answer anyways, which is my fault on my part. Thank you both for the help anyways, sorry about the confusion!

[Chunru]

The comment-out code above is good when acos become complex.