5.2 Limits and Continuity
The limit of \(f(x,y)\) as \((x,y) \longrightarrow (x_0,y_0)\) is the number \(L\) if for any \(\varepsilon > 0\), there exists a \(\delta >0\) such that for all points \((x,y) \neq (x_0,y_0)\) in the domain of \(f\), either
- 1.
- \(\sqrt {(x - x_0)^2 + (y - y_0)^2} < \delta \implies \left |f(x) - L\right | < \varepsilon \). or
- 2.
- \(\left |x - x_0\right | < \delta \) and \(\left |y - y_0\right | < \delta \implies \left |f(x) - L\right |< \varepsilon \).
If \(\lim \limits _{(x,y) \rightarrow (x_0,y_0)}f(x,y) = L_1\) and \(\lim \limits _{(x,y) \rightarrow (x_0,y_0)}g(x,y) = L_2\). Then
- 1.
- \(\lim \Big [f(x,y) \pm g(x,y)\Big ] = L_1\pm L_2\)
- 2.
- \(\lim \Big [f(x,y)\cdot g(x,y)\Big ] = L_1\cdot L_2\)
- 3.
- \(\lim \Big [Kf(x,y)\Big ] = K L_1\)
- 4.
- \(\lim \Bigg [\dfrac {f(x,y)}{g(x,y)}\Bigg ] = \dfrac {L_1}{L_2}\) if \(L_2 \neq 0\).
Evaluate the following limits:
- 1.
- \(\lim \limits _{(x,y) \rightarrow (3,-4)}\Big (x^2 + y^2\Big )\)
- 2.
- \(\lim \limits _{(x,y) \rightarrow (0,1)}\dfrac {x - xy + 3}{x^2y + 5xy - y^3}\)
Solution.
- 1.
- \(\lim \limits _{(x,y) \rightarrow (3,-4)}\Big (x^2 + y^2 \Big ) = 3^2 + (-4)^2 = 9 + 16 = 25\)
- 2.
- \(\lim \limits _{(x,y) \rightarrow (0,1)}\dfrac {x - xy + 3}{x^2y + 5xy - y^3} = \dfrac {0 - (0)(1) + 3}{(0)^2 (1) + 5(0)(1) - (1)^3} = \dfrac {3}{-1} = -3\)
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