2.2: Practice Set

Along the path \(y=1+m(x-1)\), this limit becomes:

\[ \begin{align*} \lim_{(x,y)\to (1,1)}\dfrac{y-1}{x-1} &= \lim_{x\to 1}\dfrac{(1+m(x-1))-1}{x-1}\\[4pt]
&=\lim_{x\to 1} \dfrac{m(x-1)}{x-1}=m \end{align*} \]

Since this result depends on \(m\), the limit value depends on the path taken, and does not exist.

Along the path \(x=4+m(y-2)\), this limit becomes:

\[ \begin{align*} \lim_{(x,y)\to (4,2)}\dfrac{x-y^2}{x-2y} &= \lim_{y\to 2}\dfrac{(4+m(y-2))-y^2}{(4+m(y-2)-2y} \\[4pt]
&= \lim_{y\to 2}\dfrac{(4-2m)+my-y^2}{(4-2m)+(m-2)y} \\[4pt]
&= \lim_{y\to 2}\dfrac{m-2y}{m-2} \\[4pt]
&= \dfrac{m-4}{m-2} \end{align*} \]

Since this result depends on \(m\), the limit value depends on the path taken, and does not exist.

The function \(f(x,y)=\dfrac{\cos(x)}{\cos^{-1}(y)}\) is continuous for all \(y\ne 1\), and \(f(0,0)=\dfrac{\cos(0)}{\cos^{-1}(0)}=\dfrac{1}{\frac{\pi}{2}}=\dfrac{2}{\pi}\)

Therefore we know \(\displaystyle{ \lim_{(x,y)\to (0,0)}\dfrac{\cos(x)}{\cos^{-1}(y)}}=f(0,0)=\dfrac{2}{\pi}\)

For any function \(y(x)\) satisfying \(\displaystyle \lim_{x\to 0} y(x)=0\), this limit becomes:

\[ \begin{align*} \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{x^2+y^2}} &= \lim_{x\to 0}\dfrac{y(x)+xy'(x)}{\frac{1}{2}(x^2+[y(x)]^2)^{-\frac{1}{2}}\cdot (2x+2y(x)y'(x))} \\[4pt]
&= \lim_{x\to 0}\dfrac{(y(x)+xy'(x))\sqrt{x^2+[y(x)]^2}}{x+y(x)y'(x)} \\[4pt]
&= \lim_{x\to 0}\dfrac{(y(x)+xy'(x))\cdot \frac{1}{2}(x^2+[y(x)]^2)^{-\frac{1}{2}}\cdot (2x+2y(x)y'(x))+(2y'(x)+xy''(x))\cdot \sqrt{x^2+[y(x)]^2}}{1+[y'(x)]^2+y(x)y''(x)} \\[4pt]
&= \dfrac{0}{1+[y'(0)]^2}=0 \end{align*} \]

For any function \(x(y)\) satisfying \(\displaystyle \lim_{y\to 0} x(y)=0\), this limit becomes:

\[ \begin{align*} \lim_{(x,y)\to (0,0)}\dfrac{xy}{\sqrt{x^2+y^2}} &= \lim_{y\to 0}\dfrac{x'(y)y+x(y)}{\frac{1}{2}([x(y)]^2+y^2)^{-\frac{1}{2}}\cdot (2x(y)x'(y)+2y)} \\[4pt]
&= \lim_{x\to 0}\dfrac{(x'(y)y+x(y))\sqrt{[x(y)]^2+y^2}}{x(y)x'(y)+y} \\[4pt]
&= \lim_{x\to 0}\dfrac{(x'(y)y+x(y))\cdot \frac{1}{2}([x(y)]^2+y^2)^{-\frac{1}{2}}\cdot (2x(y)x'(y)+2y)+ (2x'(y)+x''(y)y)\cdot \sqrt{[x(y)]^2+y^2}}{[x'(y)]^2+x(y)x''(y)+1} \\[4pt]
&= \dfrac{0}{1+[x'(0)]^2}=0 \end{align*} \]

Therefore all paths through the origin yields the same limit value, which is \(0\).

For any function \(y(x)\) satisfying \(\displaystyle \lim_{x\to 0} y(x)=0\), this limit becomes:

\[ \begin{align*} \lim_{(x,y)\to (0,0)}\dfrac{2x^2y^2}{x^2+4y^2} &= \lim_{x\to 0}\dfrac{2x^2[y(x)]^2}{x^2+4[y(x)]^2} \\[4pt]
&= \lim_{x\to 0}\dfrac{4x[y(x)]^2+4x^2y(x)y'(x)}{2x+8y(x)y'(x)} \\[4pt]
&= \lim_{x\to 0}\dfrac{4[y(x)]^2+16xy(x)y'(x)+4x^2[y'(x)]^2+4x^2y(x)y''(x)}{2+8[y'(x)]^2+8y(x)y''(x)} \\[4pt]
&= \dfrac{0}{2}=0 \end{align*} \]

For any function \(x(y)\) satisfying \(\displaystyle \lim_{y\to 0} x(y)=0\), this limit becomes:

\[ \begin{align*} \lim_{(x,y)\to (0,0)}\dfrac{2x^2y^2}{x^2+4y^2} &= \lim_{y\to 0}\dfrac{2[x(y)]^2y^2}{[x(y)]^2+4y^2} \\[4pt]
&= \lim_{y\to 0}\dfrac{4x(y)x'(y)y^2+4[x(y)]^2y}{2x(y)x'(y)+8y} \\[4pt]
&= \lim_{y\to 0}\dfrac{4[x'(y)]^2y^2+4x(y)x''(y)y^2+16x(y)x'(y)+4[x(y)]^2}{2[x'(y)]^2+2x(y)x''(y)+8} \\[4pt]
&= \dfrac{0}{8}=0 \end{align*} \]

Therefore all paths through the origin yields the same limit value, which is \(0\).

Along the path \(y=mx^2\), this limit becomes:

\[ \begin{align*} \lim_{(x,y)\to (0,0)}\dfrac{2x^2y}{x^4+4y^2} &= \lim_{x\to 0}\dfrac{2x^2(mx^2)}{x^4+4(mx^2)^2} \\[4pt]
&= \lim_{x\to 0}\dfrac{2mx^4}{x^4(1+4m^2)} \\[4pt]
&= \dfrac{2m}{1+4m^2} \end{align*} \]

Since this result depends on \(m\), the limit value depends on the path taken, and does not exist.

For any function \(y(x)\) satisfying \(\displaystyle \lim_{x\to 0} y(x)=0\), this limit becomes:

\[ \begin{align*} \lim_{(x,y)\to (0,0)}\dfrac{x^4-9y^4}{x^2+3y^2} &= \lim_{x\to 0}\dfrac{x^4-9[y(x)]^4}{x^2+3[y(x)]^2} \\[4pt]
&= \lim_{x\to 0}\dfrac{4x^3-36[y(x)]^3y'(x)}{2x+6y(x)y'(x)} \\[4pt]
&= \lim_{x\to 0}\dfrac{12x^2-108[y(x)]^2[y'(x)]^2-36[y(x)]^3y''(x)}{2+6[y'(x)]^2+6y(x)y''(x)} \\[4pt]
&= \dfrac{0}{2+6[y'(0)]^2}=0 \end{align*} \]

For any function \(x(y)\) satisfying \(\displaystyle \lim_{y\to 0} x(y)=0\), this limit becomes:

\[ \begin{align*} \lim_{(x,y)\to (0,0)}\dfrac{x^4-9y^4}{x^2+3y^2} &= \lim_{y\to 0}\dfrac{[x(y)]^4-9y^4}{[x(y)]^2+3y^2} \\[4pt]
&= \lim_{y\to 0}\dfrac{4[x(y)]^3x'(y) -36y^3}{2x(y)x'(y)+6y} \\[4pt]
&= \lim_{y\to 0}\dfrac{12[x(y)]^2[x'(y)]^2+4[x(y)]^3x''(y)-108y^2}{2[x'(y)]^2+2x(y)x''(y)+6} \\[4pt]
&= \dfrac{0}{2[x'(0)]^2+6}=0 \end{align*} \]

Therefore all paths through the origin yields the same limit value, which is \(0\).

Note that \((\pi ,1)\) is inside of the domain of \(f(x,y)=\dfrac{y^2\sin(x)}{x+y-1}\), and \(f(\pi,1)=\dfrac{1^2\cdot 0}{\pi+1-1}=0\).

Since this function is continuous on its domain, we know \(\displaystyle{ \lim_{(x,y)\to (\pi,1)}\dfrac{y^2\sin(x)}{x+y-1}=f(\pi,1)=0}\)

Notice that the function \(f(x,y)=\dfrac{y^2-x^2}{x^2+3xy+2y^2}=\dfrac{(y-x)(y+x)}{(x+y)(x+2y)}=\dfrac{y-x}{x+2y}\) on its domain.

Since \(g(x,y)=\dfrac{y-x}{x+2y}\) is continuous at \((2,-2)\), and \(g(2,-2)=\dfrac{-4}{-2}=2\), we know \(\displaystyle \lim_{(x,y)\to (2,-2)}g(x,y)=2\)

Therefore, we have \(\displaystyle{ \lim_{(x,y)\to (2,-2)}\dfrac{y^2-x^2}{x^2+3xy+2y^2}=\lim_{(x,y)\to (2,-2)} \dfrac{y-x}{x+2y}=2}\)

Note that the function \(f(x,y)=\dfrac{x^4-5(x+y)^2-1}{x^2+9y^2}\) is continuous everywhere, including \((3,1)\).

Since \(f(3,1)=\dfrac{0}{18}=0\), we know that \(\displaystyle{ \lim_{(x,y)\to (3,1)} \dfrac{x^4-5(x+y)^2-1}{x^2+9y^2}=0}\)

Substituting in \(a=x^4+y^4\), we know that \(a\to 0^+\) along all paths satisfying \((x,y)\to (0,0)\) so our limit becomes:

\[ \lim_{(x,y)\to (0,0)}\dfrac{\sin(x^4+y^4)}{x^4+y^4} = \lim_{a\to 0^+}\dfrac{\sin(a)}{a}= \lim_{a\to 0^+}\dfrac{\cos(a)}{1} = \cos(0)=1 \nonumber \]

Therefore this limit must be equal to \(1\).

Substituting in \(a=x+y+z\), we know that \(a\to 1\) along all paths satisfying \((x,y,z)\to \left(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}\right)\) so our limit becomes:

\[ \lim_{(x,y,z)\to \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)}\dfrac{\ln(x+y+z)}{(x+y+z)^2-1} = \lim_{a\to 1}\dfrac{\ln(a)}{a^2-1}= \lim_{a\to 1}\dfrac{\frac{1}{a}}{2a} = \lim_{a\to 1}\dfrac{1}{2a^2}= \dfrac{1}{2} \nonumber \]

Therefore this limit must be equal to \(\dfrac{1}{2}\).

Substituting in \(a=(x-1)^2+(y-2)^2+(z+1)^2\), we know that \(a\to 0^+\) along all paths satisfying \((x,y,z)\to \left(1,2,-1\right)\) so our limit becomes:

\[\lim_{(x,y,z)\to (1,2,-1)}\dfrac{1}{\sqrt{ (x-1)^2+(y-2)^2+(z+1)^2}} = \lim_{a\to 0^+}\dfrac{1}{\sqrt{a}} = +\infty \nonumber \]

Therefore all paths to \( \left(1,2,-1\right)\) yields a limit of positive infinity.