1. The aforementioned theorems allow us to simply evaluate \(y/x+\cos(xy)\) when \(x=1\) and \(y=\pi\text{.}\) If an indeterminate form is returned, we must do more work to evaluate the limit; otherwise, the result is the limit. Therefore \begin{align*} \lim_{(x,y)\to (1,\pi)} \frac yx + \cos(xy) \amp = \frac\pi{1}+\cos(\pi) \\ \amp = \pi -1. \end{align*}

2. We attempt to evaluate the limit by substituting 0 in for \(x\) and \(y\text{,}\) but the result is the indeterminate form “\(0/0\text{.}\)” To evaluate this limit, we must “do more work,” but we have not yet learned what “kind” of work to do. Therefore we cannot yet evaluate this limit.

1. Evaluating \(\lim\limits_{(x,y)\to (0,0)} \frac{3xy}{x^2+y^2}\) along the lines \(y=mx\) means replace all \(y\)'s with \(mx\) and evaluating the resulting limit: \begin{align*} \lim_{(x,mx)\to (0,0)} \frac{3x(mx)}{x^2+(mx)^2} \amp =\lim_{x\to 0} \frac{3mx^2}{x^2(m^2+1)}\\ \amp = \lim_{x\to 0} \frac{3m}{m^2+1}\\ \amp = \frac{3m}{m^2+1}. \end{align*} While the limit exists for each choice of \(m\text{,}\) we get a different limit for each choice of \(m\text{.}\) That is, along different lines we get differing limiting values, meaning the limit does not exist.

2. Let \(f(x,y) = \frac{\sin(xy)}{x+y}\text{.}\) We are to show that \(\lim\limits_{(x,y)\to (0,0)} f(x,y)\) does not exist by finding the limit along the path \(y=-\sin(x)\text{.}\) First, however, consider the limits found along the lines \(y=mx\) as done above. \begin{align*} \lim_{(x,mx)\to (0,0)} \frac{\sin\big(x(mx)\big)}{x+mx} \amp = \lim_{x\to 0} \frac{\sin(mx^2)}{x(m+1)}\\ \amp = \lim_{x\to 0} \frac{\sin(mx^2)}{x}\cdot\frac1{m+1}. \end{align*} By applying L'Hôpital's Rule, we can show this limit is 0 except when \(m=-1\text{,}\) that is, along the line \(y=-x\text{.}\) This line is not in the domain of \(f\text{,}\) so we have found the following fact: along every line \(y=mx\) in the domain of \(f\text{,}\) \(\lim\limits_{(x,y)\to(0,0)} f(x,y)=0\text{.}\) Now consider the limit along the path \(y=-\sin(x)\text{:}\) \begin{align*} \lim_{(x,-\sin(x) )\to (0,0)} \frac{\sin\big(-x\sin(x) \big)}{x-\sin(x) } \amp = \lim_{x\to0} \frac{\sin\big(-x\sin(x) \big)}{x-\sin(x) } \end{align*} Now apply L'Hôpital's Rule twice: \begin{align*} \amp = \lim_{x\to 0}\frac{\cos\big(-x\sin(x) \big)(-\sin(x) -x\cos(x) )}{1-\cos(x) } \left(\text{ `` } = 0/0\text{ '' } \right)\\ \amp = \lim_{x\to 0}\frac{-\sin\big(-x\sin(x) \big)(-\sin(x) -x\cos(x) )^2+\cos\big(-x\sin(x) \big)(-2\cos(x) +x\sin(x) )}{\sin(x) }\\ \amp = \text{ ``2/0'' } \Rightarrow \text{ the limit does not exist. } \end{align*} Step back and consider what we have just discovered. Along any line \(y=mx\) in the domain of the \(f(x,y)\text{,}\) the limit is 0. However, along the path \(y=-\sin(x)\text{,}\) which lies in the domain of \(f(x,y)\) for all \(x\neq 0\text{,}\) the limit does not exist. Since the limit is not the same along every path to \((0,0)\text{,}\) we say \(\lim\limits_{(x,y)\to (0,0)}\frac{\sin(xy)}{x+y}\) does not exist.

It is relatively easy to show that along any line \(y=mx\text{,}\) the limit is 0. This is not enough to prove that the limit exists, as demonstrated in the previous example, but it tells us that if the limit does exist then it must be 0.

To prove the limit is 0, we apply Definition 12.2.6. Let \(\varepsilon >0\) be given. We want to find \(\delta >0\) such that if \(\sqrt{(x-0)^2+(y-0)^2} \lt \delta\text{,}\) then \(\abs{f(x,y)-0} \lt \varepsilon\text{.}\)

Set \(\delta \lt \sqrt{\varepsilon/5}\text{.}\) Note that \(\ds \abs{\frac{5y^2}{x^2+y^2}} \lt 5\) for all \((x,y)\neq (0,0)\text{,}\) and that if \(\sqrt{x^2+y^2} \lt \delta\text{,}\) then \(x^2\lt \delta^2\text{.}\)

Let \(\sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2}\lt \delta\text{.}\) Consider \(\abs{f(x,y)-0}\text{:}\) \begin{align*} \abs{f(x,y)-0} \amp = \abs{\frac{5x^2y^2}{x^2+y^2}-0}\\ \amp = \abs{x^2\cdot\frac{5y^2}{x^2+y^2}}\\ \amp \lt \delta^2\cdot 5\\ \amp \lt \frac{\varepsilon}{5}\cdot 5\\ \amp = \varepsilon. \end{align*}

Thus if \(\sqrt{(x-0)^2+(y-0)^2}\lt \delta\) then \(\abs{f(x,y)-0}\lt \varepsilon\text{,}\) which is what we wanted to show. Thus \(\lim\limits_{(x,y)\to(0,0)} \frac{5x^2y^2}{x^2+y^2} = 0\text{.}\)

To determine if \(f\) is continuous at \((0,0)\text{,}\) we need to compare \(\lim\limits_{(x,y)\to (0,0)} f(x,y)\) to \(f(0,0)\text{.}\)

Applying the definition of \(f\text{,}\) we see that \(f(0,0) = \cos(0) = 1\text{.}\)

We now consider the limit \(\lim\limits_{(x,y)\to (0,0)} f(x,y)\text{.}\) Substituting \(0\) for \(x\) and \(y\) in \((\cos(y) \sin(x) )/x\) returns the indeterminate form “0/0”, so we need to do more work to evaluate this limit.

Consider two related limits: \(\lim\limits_{(x,y)\to (0,0)} \cos(y)\) and \(\lim\limits_{(x,y)\to(0,0)} \frac{\sin(x) }x\text{.}\) The first limit does not contain \(x\text{,}\) and since \(\cos(y)\) is continuous, \begin{equation*} \lim\limits_{(x,y)\to (0,0)} \cos(y) =\lim_{y\to 0} \cos(y) = \cos(0) = 1. \end{equation*}

The second limit does not contain \(y\text{.}\) By Theorem 1.3.12 we can say \begin{equation*} \lim_{(x,y)\to (0,0)} \frac{\sin(x) }{x} = \lim_{x\to 0} \frac{\sin(x) }{x} = 1. \end{equation*}

Finally, Theorem 12.2.8 of this section states that we can combine these two limits as follows: \begin{align*} \lim_{(x,y)\to (0,0)} \frac{\cos(y) \sin(x) }{x} \amp = \lim_{(x,y)\to (0,0)} (\cos(y) )\left(\frac{\sin(x) }{x}\right)\\ \amp =\left(\lim_{(x,y)\to (0,0)} \cos(y) \right)\left(\lim_{(x,y)\to (0,0)} \frac{\sin(x) }{x}\right)\\ \amp = (1)(1)\\ \amp =1. \end{align*}

We have found that \(\lim\limits_{(x,y)\to (0,0)} \frac{\cos(y) \sin(x) }{x} = f(0,0)\text{,}\) so \(f\) is continuous at \((0,0)\text{.}\)

A similar analysis shows that \(f\) is continuous at all points in \(\mathbb{R}^2\text{.}\) As long as \(x\neq0\text{,}\) we can evaluate the limit directly; when \(x=0\text{,}\) a similar analysis shows that the limit is \(\cos(y)\text{.}\) Thus we can say that \(f\) is continuous everywhere. A graph of \(f\) is given in Figure 12.2.14. Notice how it has no breaks, jumps, etc.

We will apply both Theorems 1.5.8 and 12.2.15. Let \(f_1(x,y) = x^2\text{.}\) Since \(y\) is not actually used in the function, and polynomials are continuous (by Theorem 1.5.8), we conclude \(f_1\) is continuous everywhere. A similar statement can be made about \(f_2(x,y) = \cos(y)\text{.}\) Part 3 of Theorem 12.2.15 states that \(f_3=f_1\cdot f_2\) is continuous everywhere, and Part 7 of the theorem states the composition of sine with \(f_3\) is continuous: that is, \(\sin(f_3) = \sin(x^2\cos(y) )\) is continuous everywhere.