I am stuck with this problem and do not know how to proceed. I have been dealing with Discrete Math for years and forgot many things about Calculus.
Consider the function $f:\mathbb{R}\setminus \{(0,0)\}\to \mathbb{R}$ such that $$ f(x,y) = \frac{e^{x+y}-x-y-1}{x^2+y^2}. $$ Determine $f(0,0)$ such that $f'_x(0,0)$ exists and also determine $f'_x(0,0)$.
What I have tried:
We know that $$ f'_x(0,0) = \lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h}. $$ Firstly I tried to calculate: $$ \lim_{h\to 0} \frac{f(0+h,0)}{h} = \lim_{h\to 0} \frac{e^h-h-1}{h^3} $$ but the limit does not exist. So I do not think that it is possible to give any value to $f(0,0)$ such that $f'_x(0,0)$ exists. Am I missing something? Any help is appreciated.
You seem to have assumed that $ f ( 0 , 0 ) $ is $ 0 $, and that didn't work. Since you don't know what $ f ( 0 , 0 ) $ should be, just write it as $ c $ or something. So you have $$ \lim _ { h \to 0 } \frac { f ( 0 + h , 0 ) - f ( 0 , 0 ) } h = \lim _ { h \to 0 } \frac { \frac { \mathrm e ^ h - h - 1 } { h ^ 2 } - c } h \text . $$ Since the limit of the denominator is $ 0 $, the only way that this limit can exist is if the limit of the numerator is also $ 0 $; that will tell you what $ c $ must be. This doesn't guarantee that the limit exists even with that value of $ c $, but you can try it, and it turns out that it does exist, and you can see what it is.
Another way to approach this is to use the theorem that a function can only be differentiable when it's continuous. We're not talking here about the continuity and differentiability of $ f $ itself (since we're only looking at one partial derivative) but of the function $ x \mapsto f ( x , 0 ) $. This function of one variable is differentiable at $ 0 $ iff $ f ' _ x ( 0 , 0 ) $ exists, and so (if this is true) we can find its value at $ 0 $ by the limit $ f ( 0 , 0 ) = \lim \limits _ { x \to 0 } f ( x , 0 ) $. This will be the same as the value that you find for $ c $ by the first method. Then you still have to go back to the definition of the derivative as a limit to find the value of $ f ' _ x ( 0 , 0 ) $ (which still might not exist, although in this case it does).