236CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATIONbut if we try to calculated(0,0)f(−→v) for the vector−→v= (1,m) usingEquation (3.6),d(0,0)f(1,m) = limt→01t(f(t,mt)−f(0,0))then, since along the liney=mxthe function has a constant valuef(t,mt) =m1 +m2—which is nonzero ifmis—butf(0,0) = 0, we see that the limit abovedoes not exist:limt→01t(f(t,mt)−f(0,0)) = limt→01tparenleftbiggm1 +m2parenrightbiggdiverges, and the differential cannot be evaluated along the vector−→v=−→ı+m−→ifmnegationslash= 0. In fact, we saw before that this function is notcontinuous at the origin, which already contradicts differentiability, byRemark3.3.2.Another example, this time one whichiscontinuous at the origin, isf(x,y) =2xy√x2+y2if (x,y)negationslash= (0,0),0at (0,0).This function is better understood when expressed in polar coordinates,
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