I don't know what you mean by "t+> 0".
To find the directional derivative of f(x,y), in the direction that makes angle [itex]\theta[/itex] with the xaxis, you can use the unit vector [itex]cos(\theta)\vec{i}+ sin(\theta)\vec{j}[/itex].
In particular, for f(x,y)= x+ y, at (0, 0) we have
[tex]D_\theta f(0, 0)= \lim_{t\to 0}\frac{f(tcos(\theta), tsin(\theta)}{t}[/tex]
[tex]= \lim_{t\to 0}\frac{tcos(\theta)+ tsin(\theta)}{t}= (cos(\theta)+ sin(\theta)\lim_{t\to 0}\frac{2t}{t}[/tex]
and that is clearly going to give different results for t approaching 0 from above or below.
