Question

h(x,y)=(cos(x)-cos(y))/(x2+y2). Can this function have a value c so that c=h(0,0)?

Answer 1

I thought about this solution as a kinda trick, just don't know if it is legally or mathematically correct:
\[f(x)=h(x,0)=\frac{ \cos(x)-1 }{ x^2 }\]
Replacing cos (x) with the Taylor development:
\[\cos(x)=1-\frac{ x^2 }{ 2 }+x^2 * e(x)\]
Here, e (x) is a continuous function with e (x) -> 0 for x -> 0. It provides by substitution in f (x):
\[f(x)=\frac{ (1-\frac{ x^2}{ 2 }+x^2 * e(x))-1 }{ x^2 }=\frac{ -\frac{ x^2 }{ 2 }+x^2 * e(x) }{ x^2 }\]
\[=\frac{ -1 }{ 2 }+e(x)\]
\[\lim_{x \rightarrow 0}(\frac{ -1 }{ 2 }+e(x))=\frac{ -1 }{ 2 }\]

This means that, if h(x, y) must be continuous across R^2 there must apply that h(0,0)= -1/2
Therefore, we now look opon the following: h(0,0)=-1/2.

By replacing cos(y) with the Taylor development it can be seen that h(0, y)->1/2 for y->0 The conclusion is that h (x, y) is continuous in the (0.0) as
h(0, y) -> 1/2 ≠ h(0,0) for y -> 0

Answer 2

Seems reasonable, although the way the question is phrased, I would have said undefined.

I think the question should say something along the lines of "Does the limit exist at the origin", that would be clearer.