Answer to Question #289438 in Calculus for Pankaj

Question

Let f(x, y) ={y³/x²+y² if (x, y) ≠(0, 0)

0 otherwise

Show that f is continuous but not differentiable at (0, 0)

Answer 1

$\text{By definition, }f(0,0)=0.\\ \text{Now consider the limits along straight lines x=t, y=at(or y=ax, where a is the Slope) as}\\ t\rightarrow0^+. \\ \text{Then,}\\ \lim_{t\rightarrow0^+}f(t,at)=\lim_{t\rightarrow0^+}\frac{(at)^3)}{t^2+(at)^2}=\lim_{t\rightarrow0^+}\frac{a^3t^3}{t^2+a^2t^2}=\lim_{t\rightarrow0^+}\frac{a^3t}{1+a^2}=0\\ \quad\\ \text{Next, If the two-variable limit exists, then it must be equal to }0.\\ \text{This can be verified by the means of the simplified squeeze principle; that is, one has to}\\ \text{ verify that }\exist\ h(R) \ni |f(x,y)-f(0,0)|\leq h(R)\rightarrow0 \text{ as} \ R=\sqrt{x^2+y^2}\rightarrow0.\\ \text{Now,}\\ \quad\\ |f(x,y)-f(0,0)|=|\frac{y^3}{x^2+y^2}|\leq\frac{|y^3|}{|x^2+y^2|}\leq\frac{R^3}{2R^2}=\frac{R}{2}\rightarrow0, \text{as }R\rightarrow0 \text{ where the inequalities}\\ |x|\leq R \text{ and }|y|\leq R \text{ have been used . Thus, the two-variable limit exists at (0,0) and equals 0=f(0,0).}\\ \text{We can therefore conclude that the given function is continuous.}\\ \quad\\ \text{Differentiating with respect to x yields: }f_x(x,y)=\frac{-2xy^3}{(x^2+y^2)^2}\\ \quad\text{Now, }f_x(0,0)=0, \text{but}\\ \text{Now consider the limits along straight lines x=t, y=at(or y=ax, where a is the Slope) as}\\ t\rightarrow0^+. \\ \qquad\text{Then,}\lim_{(x,y)\rightarrow(0,0)}f_x(x,y)=\lim_{(x,y)\rightarrow(0,0)}\frac{-2xy^3}{(x^2+y^2)^2}=\lim_{t\ \rightarrow\ 0^+}\frac{-2t(at)^3}{(t^2+(at)^2)^2}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ \ \ \ =\lim_{t\ \rightarrow\ 0^+}\frac{-2a}{(1+a^2)^2}=\frac{-2a}{(1+a^2)^2}.\\ \text{Since the limit along a straight line depends on the slope of the line. Therefore, the}\\ \text{ two-variable limit does not exist at (0, 0). Since the limit does not exist, this shows that }\\ \text{the partial derivative of the function with respect to x exist but is not continuous.}\\ \text{This shows that the function is not differentiable}.$