## Calculus: Early Transcendentals 8th Edition

We notice that if we directly substitute limits in the given function $f(x,y)=\frac{y^{2}sin^{2}x}{x^{4}+y^{4}}$ Then $f(0,0)=\frac{0}{0}$ Therefore, we will calculate limit of function in following way. To evaluate limit along x-axis; put $y=0$ $f(x,0)=\frac{y^{2}sin^{2}x}{x^{4}+y^{4}}=\frac{(0)^{2}sin^{2}x}{x^{4}+0}=0$ To evaluate limit along y-axis; put $x=0$ $f(0,y)=\frac{y^{2}sin^{2}0}{(0)^{4}+y^{4}}=0$ However, the obtained identical limits along the axes do not show that the given limit is 0. Then, approach (0,0) along another line, $y=x$ for $x\ne0$ $f(x,x)=\frac{x^{2}sin^{2}x}{x^{4}+x^{4}}=\frac{sin^{2}x}{2x^{4}}$ $f(x,x)=\frac{1}{2}(\frac{sinx}{x})^{2}$ Now, $\lim\limits_{(x,y) \to (0,0)}f(x,x)=\lim\limits_{(x,y) \to (0,0)}\frac{1}{2}(\frac{sinx}{x})^{2}$ $\lim\limits_{(x,y) \to (0,0)}f(x,x)=\frac{1}{2}$ For a limit to exist, all the paths must converge to the same point. Hence, both the limits are different and follow different paths, therefore, the limits does not exist.