## Calculus 8th Edition

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We notice that if we directly substitute limits in the given function $f(x,y)=\frac{xy}{\sqrt {x^{2}+y^{2}}}$ Then $f(0,0)=\frac{0}{0}$ We get intermediate form. Therefore, we will calculate limit of function in following way. To evaluate limit along x-axis; put $y=0$ $f(x,0)=\frac{x.0}{\sqrt {x^{2}+0^{2}}}=0$ To evaluate limit along y-axis; put $x=0$ $f(0,y)=\frac{0.y}{\sqrt {0^{2}+y^{2}}}=0$ Although the obtained identical limits along the axes do not show that the given limit is 0. Then, approach (0,0) along another line let us say $y^{2}=x$ and $x\ne0$ $f(y^{2},y)=\frac{0.y}{\sqrt {0^{2}+y^{2}}}=\frac{y^{2}}{\sqrt {y^{2}+1}}$ $\lim\limits_{(x,y) \to (0,0)}f(y^{2},y)=\lim\limits_{(x,y) \to (0,0)}\frac{y^{2}}{\sqrt {y^{2}+1}}=0$ Then, approach (0,0) along another line let us say $x^{2}=y$ and $x\ne0$ $f(x,x^{2})=\frac{0.y}{\sqrt {0^{2}+y^{2}}}=\frac{x^{2}}{\sqrt {x^{2}+1}}$ $\lim\limits_{(x,y) \to (0,0)}f(x,x^{2})=\lim\limits_{(x,y) \to (0,0)}\frac{x^{2}}{\sqrt {x^{2}+1}}=0$ Hence, limit =0.