Using Properties of Limits to Find a Limit

Find each limit:

1. \(\lim\limits_{(x, y)\rightarrow (1, 2)}(3x^{2}+2xy+y^{2})\)
2. \(\lim\limits_{(x, y)\rightarrow (1, 2)}\dfrac{xy}{x^{2}+y^{2}}\)
3. \(\lim\limits_{(x, y)\rightarrow (3, -1)}\dfrac{x^{2}+2xy-3y^{2}}{ x^{2}+9y^{2}}\)

Solution (a) \[ \begin{eqnarray*} \lim_{(x, y)\rightarrow (1, 2)}(3x^{2}+2xy+y^{2})& =&\lim_{(x, y)\rightarrow (1, 2)}( 3x^{2}) +\lim_{(x, y)\rightarrow (1, 2)}( 2xy) +\lim_{(x, y)\rightarrow (1, 2)}y^{2} \\[8pt] & =&3\lim_{x\rightarrow 1}x^{2}+\Big( \lim_{(x, y)\rightarrow (1, 2)}( 2x) \Big) \Big( \lim_{(x, y)\rightarrow (1, 2)}y\Big) +\lim_{y\rightarrow 2}y^{2} \\[8pt] &=& 3+2\Big( \lim_{x\rightarrow 1}x\Big) \Big( \lim_{y\rightarrow 2}y\Big) +4=3+2\cdot 1\cdot 2+4=11 \end{eqnarray*} \]

(b) Since \(\lim\limits_{(x, y)\rightarrow (1, 2)} (x^2+y^2) = \lim\limits_{x\rightarrow 1} x^2 + \lim\limits_{y\rightarrow 2} y^2 =5\neq 0\), we use the limit of a quotient: \[ \begin{eqnarray*} \lim_{(x, y)\rightarrow (1, 2)}\frac{xy}{x^{2}+y^{2}} &=&\frac{ \lim\limits_{(x, y)\rightarrow (1, 2)}(xy) }{\lim\limits_{(x, y) \rightarrow (1, 2)}(x^{2}+y^{2})} =\frac{\Big( \lim\limits_{x\rightarrow 1}x\Big) \Big( \lim\limits_{y\rightarrow 2}y\Big) }{\lim\limits_{x\rightarrow 1}x^{2}+\lim\limits_{y\rightarrow 2}y^{2}}=\frac{1\cdot 2}{1+4}=\frac{2}{5} \end{eqnarray*} \]

(c) Look at the denominator. Since \(\lim\limits_{(x, y) \rightarrow (3,-1)}( x^{2}+9y^{2}) =18,\) we can use the limit of a quotient property. Then \[ \begin{eqnarray*} \lim\limits_{(x, y)\rightarrow (3,-1)}\dfrac{x^{2}+2xy-3y^{2}}{x^{2}+9y^{2}} &=&\dfrac{\lim\limits_{(x, y)\rightarrow (3,-1)}(x^{2} + 2xy - 3y^{2}) }{\lim\limits_{(x, y)\rightarrow (3,-1)}(x^{2} + 9y^{2}) }\\[8pt] &=&\frac{\lim\limits_{x\rightarrow 3}x^{2} + 2\Big( \lim\limits_{x\rightarrow 3}x\Big) \Big( \lim\limits_{y\rightarrow -1}y\Big) - 3 \lim\limits_{y\rightarrow -1}y^{2}}{18}\\[8pt] &=&\dfrac{9 + 2(3)(-1) - 3(1)}{18}=0 \end{eqnarray*} \]