1. First use the sum and difference laws to separate the terms:

[latex]\hspace{5cm}\displaystyle\lim_{(x,\ y)\to(2,\ -1)}(x^{2}-2xy+3y^{2}-4x+3y-6)[/latex]

Next, use the constant multiple law on the second, third, fourth, and fifth limits:

Now, use the power law on the first and third limits, and the product law on the second limit:

Last, use the identity laws on the first six limits and the constant law on the last limit:

1. Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Using the difference law, constant multiple law, and identity law,
   [latex]\begin{array}{ccc}\hfill {\displaystyle\lim_{(x,\ y)\to(2,\ -1)}(4x-3y)} & =\hfill & {\displaystyle\lim_{(x,\ y)\to(2,\ -1)}4x-\displaystyle\lim_{(x,\ y)\to(2,\ -1)}3y}\hfill \\ \hfill & =\hfill & {4\big(\displaystyle\lim_{(x,\ y)\to(2,\ -1)}x\big)-3\big(\displaystyle\lim_{(x,\ y)\to(2,\ -1)}y\big)} \hfill \\ \hfill & =\hfill & {4(2)-3(-1)=11.}\hfill \\ \hfill \end{array}[/latex]

    

   Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of the numerator using the difference law, constant multiple law, and identity law:

   [latex]\begin{array}{ccc}\hfill {\displaystyle\lim_{(x,\ y)\to(2,\ -1)}(2x+3y)} & =\hfill & {\displaystyle\lim_{(x,\ y)\to(2,\ -1)}2x+\displaystyle\lim_{(x,\ y)\to(2,\ -1)}3y}\hfill \\ \hfill & =\hfill & {2\big(\displaystyle\lim_{(x,\ y)\to(2,\ -1)}x\big)+3\big(\displaystyle\lim_{(x,\ y)\to(2,\ -1)}y\big)} \hfill \\ \hfill & =\hfill & {2(2)+3(-1)}\hfill \\ \hfill & =\hfill & {1.}\hfill \\ \hfill \end{array}[/latex]

    

   Therefore, according to the quotient law we have

   [latex]\displaystyle\lim_{(x,\ y)\to(2,\ -1)}\frac{2x+3y}{4x-3y}=\frac{\displaystyle\lim_{(x,\ y)\to(2,\ -1)}(2x+3y)}{\displaystyle\lim_{(x,\ y)\to(2,\ -1)}(4x-3y)}=\frac{1}{11}.[/latex]

1. The domain of the function [latex]f\,(x,\ y)=\frac{2xy}{3x^{2}+y^{2}}[/latex] consists of all points in the [latex]xy[/latex]-plane except for the point [latex](0,\ 0)[/latex] (Figure 3). To show that the limit does not exist as [latex](x,\ y)[/latex] approaches [latex](0,\ 0)[/latex], we note that it is impossible to satisfy the definition of a limit of a function of two variables because of the fact that the function takes different values along different lines passing through point [latex](0,\ 0)[/latex]. First, consider the line [latex]y=0[/latex] in the [latex]xy[/latex]-plane. Substituting [latex]y=0[/latex] into [latex]f\,(x,\ y)[/latex] gives
   [latex]\large{f\,(x,\ 0)=\frac{2x(0)}{3x^{2}+0^{2}}=0}[/latex]

    

   For any value of [latex]x[/latex]. Therefore the value of [latex]f[/latex] remains constant for any point on the [latex]x[/latex]-axis, and as [latex]y[/latex] approaches zero, the function remains fixed at zero. Next, consider the line [latex]y=x[/latex]. Substituting [latex]y=x[/latex] into [latex]f\,(x,\ y)[/latex] gives

   [latex]\large{f\,(x,\ x)=\frac{2x(x)}{3x^{2}+x^{2}}=\frac{2x^{2}}{4x^{2}}=\frac{1}{2}}.[/latex]

    

   This is true for any point on the line [latex]y=x[/latex]. If we let [latex]x[/latex] approach zero while staying on this line, the value of the function remains fixed at [latex]\frac{1}{2}[/latex], regardless of how small [latex]x[/latex] is. Choose a value for [latex]\epsilon[/latex] that is less than [latex]1/2-[/latex]say, [latex]1/4[/latex]. Then, no matter how small a [latex]\delta[/latex] disk we draw around [latex](0,\ 0)[/latex], the values of [latex]f\,(x,\ y)[/latex] for points inside that [latex]\delta[/latex] disk will include both [latex]0[/latex] and [latex]\frac{1}{2}.[/latex] Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.

   

   Figure 3. Graph of the function [latex]\small{f(x,y)=(2xy)/(3x^{2}+y^{2})}[/latex]. Along the line [latex]\small{y=0}[/latex], the function is equal to zero; along the line [latex]\small{y=x}[/latex], the function is equal to [latex]\small{\frac{1}{2}}[/latex].

2. In a similar fashion to 1., we can approach the origin along any straight line passing through the origin. If we try the [latex]x[/latex]-axis (i.e., [latex]y=0[/latex]), then the function remains fixed at zero. The same is true for the [latex]y[/latex]-axis. Suppose we approach the origin along a straight line of slope [latex]k[/latex] The equation of this line is [latex]y=kx[/latex]. Then the limit becomes
   [latex]\begin{array}{ccc}\hfill {\displaystyle\lim_{(x,\ y)\to(0,\ 0)}\frac{4xy^{2}}{x^{2}+3y^{4}}} & =\hfill & {\displaystyle\lim_{(x,\ y)\to(0,\ 0)}\frac{4x(kx)^{2}}{x^{2}+3(kx)^{4}}}\hfill \\ \hfill & =\hfill & {\displaystyle\lim_{(x,\ y)\to(0,\ 0)}\frac{4k^{2}x^{3}}{x^{2}+3k^{4}x^{4}}} \hfill \\ \hfill & =\hfill & {\displaystyle\lim_{(x,\ y)\to(0,\ 0)}\frac{4k^{2}x}{1+3k^{4}x^{2}}}\hfill \\ \hfill & =\hfill & {\frac{\displaystyle\lim_{(x,\ y)\to(0,\ 0)}4k^{2}x}{\displaystyle\lim_{(x,\ y)\to(0,\ 0)}(1+3k^{4}x^{2})}}\hfill \\ \hfill & =\hfill & {0}\hfill \\ \hfill \end{array}[/latex]

    

   Regardless of the value of [latex]k[/latex]. It would seem that the limit is equal to zero. What if we chose a curve passing through the origin instead? For example, we can consider the parabola given by the equation [latex]x=y^{2}[/latex]. Substituting [latex]y^{2}[/latex] in place of [latex]x[/latex] in [latex]f\,(x,\ y)[/latex] gives

   [latex]\begin{array}{ccc}\hfill {\displaystyle\lim_{(x,\ y)\to(0,\ 0)}\frac{4xy^{2}}{x^{2}+3y^{4}}} & =\hfill & {\displaystyle\lim_{(x,\ y)\to(0,\ 0)}\frac{4(y^{2})y^{2}}{(y^{2})^{2}+3y^{4}}}\hfill \\ \hfill & =\hfill & {\displaystyle\lim_{(x,\ y)\to(0,\ 0)}\frac{4y^{4}}{y^{4}+3y^{4}}} \hfill \\ \hfill & =\hfill & {\displaystyle\lim_{(x,\ y)\to(0,\ 0)}1}\hfill \\ \hfill & =\hfill & {1.}\hfill \\ \hfill \end{array}[/latex]

    

   By the same logic in 1., it is impossible to find a [latex]\delta[/latex] disk around the origin that satisfies the definition of the limit for any value of [latex]\epsilon\,<\,1[/latex]. Therefore, [latex]\displaystyle\lim_{(x,\ y)\to(0,\ 0)}\frac{4xy^{2}}{x^{2}+3y^{4}}[/latex] does not exist.