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HallsofIvy is offline
Oct9-10, 08:24 AM
Sci Advisor
PF Gold
P: 38,879
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 x-axis, 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{|t||cos(\theta)|+ |t||sin(\theta)|}{t}= (cos(\theta)+ sin(\theta)\lim_{t\to 0}\frac{2|t||}{t}[/tex]
and that is clearly going to give different results for t approaching 0 from above or below.