14. Show that the function. Solution To show that f is not continuous at (0,0) w

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To show that f is not continuous at (0,0) we calculate the limit as(x,y) -> (0,0) and show that it is path dependent and thereforethe limit does not exist. Consider the path y=0, so lim x-> 0 of f(x,y) = 0. But if we consider the path y=x^3 thenwe get lim (x,y)->0 of f(x,y) equals lim x ->0 of x^5/3x^6 =lim x->0 of 1/x -> infinity which does not equal 0 thus thelimit does not exist. You can also show this using the substitution x=rcos,y=rsin and (x,y) -> (0,0) is equivalent to r->0 thenshow that lim r->0 f(r,) depends on (trajectorydependent) and therefore does not exist as the limit takesdifferent values for different values of . . To show that the directional derivative exists consider somearbitrary nonzero unit vector (h,k). NowD(h,k)f(0,0) = lim t ->0 of f(0+th,0+tk) - f(0,0) /t. But f(0,0) = 0 and so we have lim t ->0 of (th)^2*(tk)/((th)^6 + 2(tk)^2) * 1/t, simplifyingwe get lim t->0 of (t^3*h^2*k) / (t^7*h^6 + 2t^3*k^2) or lim t->0 (h^2*k)/ (t^4*h^6 + 2k^2) or h^2*k / 2k^2 = h/2k

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