## Thomas' Calculus 13th Edition

Consider our first approach : $(x,y) \to (1,-1)$ along $x=1$ Then, we get $\lim\limits_{y \to -1}\dfrac{y+1}{1-y^2}=\lim\limits_{y \to -1}\dfrac{(y+1)}{(1+y)(1-y)}$ or, $\dfrac{1}{1-(-1)}=\dfrac{1}{2}$ Next, consider our second approach : $(x,y) \to (1,-1)$ along $y=-1$ This implies thta $\lim\limits_{y \to -1}\dfrac{-x+1}{x^2-1}=\lim\limits_{y \to -1}\dfrac{(1-x)}{(x+1)(x-1)}$ or, $\dfrac{-1}{1+1}=-\dfrac{1}{2}$ This shows that there are different limit values when the approach is different, so, the limit does not exist at the point (1,-1)for the function $f(x,y)=\dfrac{xy+1}{x^2-y^2}$ .
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