(a)
$$
\lim_{(x,y)\to (1,\frac{\pi}{6})} \frac{x\sin y}{x^{2}+1}
$$
Solution (a):
\begin{align*}
\lim_{(x,y)\to (1,\pi/6)} \frac{x\sin y}{x^{2}+1} = \frac{1\cdot \sin \frac{\pi}{6}}{1^{2}+1}=\frac{1}{4}
\end{align*}
(b)
$$
\lim_{(x,y)\to (2,2)} \frac{x+y-4}{\sqrt{x+y}-2}
$$
Solution (b):
\begin{align*}
\lim_{(x,y)\to (2,2)} \frac{x+y-4}{\sqrt{x+y}-2} = \lim_{(x,y)\to (2,2)} \frac{(\sqrt{x+y})^{2}-2^{2}}{\sqrt{x+y}-2}=\\
\lim_{(x,y)\to (2,2)} \frac{(\sqrt{x+y}-2)(\sqrt{x+y}+2)}{\sqrt{x+y}-2}=\lim_{(x,y)\to (2,2)} (\sqrt{x+y}+2)=4
\end{align*}
Let $f(x,t)=\ln(3x+2t)$. Find all possible constants $c$ such that the function $f(x,t)$
is a solution of the wave equation:
$$
f''_{tt}-c^{2}f''_{xx}=0
$$
Solution:
\begin{align*}
f&=\ln(3x+2t)\\
f'_{x}&=\frac{3}{3x+2t}\\
f''_{xx}&=\frac{-9}{(3x+2t)^{2}}\\
\end{align*}
\begin{align*}
f&=\ln(3x+2t)\\
f'_{t}&=\frac{2}{3x+2t}\\
f''_{tt}&=\frac{-4}{(3x+2t)^{2}}\\
\end{align*}
\begin{align*}
&f''_{tt}-c^{2}f''_{xx}=\frac{-4}{(3x+2t)^{2}} +c^{2}\cdot \frac{9}{(3x+2t)^{2}}=0\\
&-4+9c^{2}=0\\
&c = \pm \frac{2}{3}
\end{align*}