202104221542 Homework 2 (Q4)

Evaluate the limit, or explain why the limit fails to exist.

(a) $\displaystyle{\lim_{(x,y)\to (0,0),\, x\neq y} \frac{x^2-xy}{\sqrt{x}-\sqrt{y}}}$ ;

(b) $\displaystyle{\lim_{(x,y)\to (2,0)}\frac{x^2-y^2-4x+4}{x^2+y^2-4x+4}}$

Solution.

(a)

$\begin{aligned} & \quad \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x^2-xy}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x(x-y)}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} x(\sqrt{x}+\sqrt{y}) \\ & = (0)\big( \sqrt{(0)} + \sqrt{(0)} \big) \\ & = 0 \end{aligned}$

(b)

Roughwork.

If we take limits along the path $(2,y)\to (2,0)$ ,

$\begin{aligned} & \quad \lim_{(x,y)\to (2,0)}\frac{x^2-y^2-4x+4}{x^2+y^2-4x+4} \\ & = \lim_{(x,y)\to (2,0)} \frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{\big((2)-2\big)^2-y^2}{\big((2)-2\big)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{-y^2}{y^2} \\ & = -1 \end{aligned}$

whereas if we take limits along the path $(x,0)\to (2,0)$ ,

$\begin{aligned} & \quad \lim_{(x,y)\to (2,0)} \frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{x\to 2\textrm{ along }y=0}\frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{x\to 2\textrm{ along }y=0}\frac{(x-2)^2-(0)^2}{(x-2)^2+(0)^2} \\ & = 1 \end{aligned}$

The limit fails to exist because the limiting values vary with the paths of taking the limit $\lim_{(x,y)\to (2,0)}$ .

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