Solution.

We want to show that \(\left |f(x,y) - f(0,0)\right | < \varepsilon \) if \(\sqrt {(x - 0)^2 + (y - 0)^2} <\delta \) or \(\sqrt {x^2 + y^2}< \delta \), Since \(f(0,0) = 0\) then \[\left |f(x,y) - f(0,0)\right | = \left |\dfrac {x + y}{x^2 + y^2 + 1}\right |\]

We want \(\left |\dfrac {x + y}{x^2 + y^2 + 1}\right |\)

Since \(x^2 + y^2 + 1 \geq 1 \implies \dfrac {1}{x^2 + y^2 + 1}\leq 1\)

\[\implies \hspace {0.4cm} \left |\dfrac {x + y}{x^2 + y^2 + 1}\right | = \dfrac {1}{x^2 + y^2 + 1}\hspace {0.1cm}\left |x + y\right |\leq \left |x + y\right |\leq \left |x\right | + \left |y\right |\]

But \(\left |x\right | = \sqrt {x^2}\) and \(\left |y\right | = \sqrt {y^2}\)

\(\implies \hspace {0.3cm} \left |x\right | = \sqrt {x^2} \leq \sqrt {x^2 + y^2}\) and \(\left |y\right |\leq \sqrt {x^2 + y^2} \hspace {0.3cm}\implies \hspace {0.3cm} \delta = \dfrac {\varepsilon }{2}\)

\begin {align*} \left |\dfrac {x + y}{x^2 + y^2 + 1}\right | & \leq \left |x\right | + \left |y\right |\\ & \leq \sqrt {x^2 + y^2} + \sqrt {x^2 + y^2}\\ & = 2\sqrt {x^2 + y^2}\\ & < 2\delta \\ & = \varepsilon \end {align*}

Therefore, the \(\varepsilon \) inequality will hold if the distance from \((x,y)\) to \((0,0)\) is less that \(\dfrac {\varepsilon }{2}\).

Solution.

Along \(y = mx, \hspace {0.3cm} f(x,y) = f(x,mx) = \dfrac {x^2 (mx)}{x^4 + (mx)^2}\hspace {0.2cm}\) so \(\hspace {0.2cm} h(x,y) = \dfrac {mx^3}{x^2(x^2 + m^2)} = \dfrac {mx}{x^2 + m^2}\) and so \[\lim \limits _{(x,y) \rightarrow (0,0)}f(x,y) = \lim \limits _{x \rightarrow 0} h(x) = \dfrac {mx}{x^2 + m^2} = 0 \hspace {0.3cm}\text {along}\hspace {0.2cm} y = mx\]

Along \(y = x^2, \hspace {0.3cm} f(x,y) = f(x,x^2) = \dfrac {x^4}{x^4 + x^4}\)

\[f(x,x^2) = \dfrac {x^4}{2x^4} = \dfrac {1}{2}\longrightarrow \dfrac {1}{2} \hspace {0.3cm} \text {as}\hspace {0.3cm} x \longrightarrow 0\]

\(\implies \hspace {0.4cm}\lim \limits _{(x,y) \rightarrow (0,0)}f(x,y) = \dfrac {1}{2}\hspace {0.3cm}\) along \(y = x^2\)

\(f\) does not have a limit as \((x,y) \longrightarrow (0,0)\) since along two different paths the limits of \(f\) as \((x,y) \longrightarrow (0,0)\) are different.

Solution.

1.

\(f(0,0) = 0 \) exists

2.

\(\lim \limits _{(x,y) \rightarrow (0,0)}f(x,y) = \lim \limits _{(x,y) \rightarrow (0,0)}\dfrac {xy}{x^2 + y^2}\)

along \(y =x\) \(g(x,y) = g(x,x) = \dfrac {x^2}{2x^2} = \dfrac {1}{2}\)

along \(y = -x\) \(g(x,y) = g(x,-x) = \dfrac {-x^2}{2x^2} = \dfrac {-1}{2}\neq \dfrac {1}{2}\)

\(\therefore \hspace {0.3cm} \lim \limits _{(x,y) \rightarrow (0,0)}f(x,y)\) does not exist.

\(\therefore \hspace {0.3cm} f(x,y)\) is not continuous at \((0,0)\).