Calculus – Functions of several variables – Limits and continuity

Definition: Let \(f\) be a function of two variables whose domain \(D\) includes points arbitrarily close to \((a,b)\). Then

\[\lim\limits_{(x,y)\to(a,b)}f(x,y)=L,\]

if for every \(\epsilon > 0\) there is a \(\delta > 0\) such that

\[|f(x,y)-L| < \epsilon\]

for all \((x,y)\in D\) with \(0 < \sqrt{(x-a)^2+(y-b)^2} < \delta\).

If

\[f(x,y)\to L_1\quad\text{as}\quad (x,y)\to(a,b)\quad\text{along a path}\quad C_1\]

and

\[f(x,y)\to L_2\quad\text{as}\quad (x,y)\to(a,b)\quad\text{along a path}\quad C_2\]

with \(L_1\neq L_2\), then \(\displaystyle\lim\limits_{(x,y)\to(a,b)}f(x,y)\) does not exist.

Examples

1) \(\displaystyle\lim\limits_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}\) does not exist. Let \(f(x,y)=\displaystyle\frac{x^2-y^2}{x^2+y^2}\). Then we have:

\[\lim\limits_{x\to0}f(x,0)=\lim\limits_{x\to0}\frac{x^2-0}{x^2+0}=1\quad\text{and}\quad\lim\limits_{y\to0}f(0,y)=\lim\limits_{y\to0}\frac{0-y^2}{0+y^2}=-1.\]

2) \(\displaystyle\lim\limits_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}\) does not exist. For \(y=0\) we have \(\displaystyle\lim\limits_{x\to0}\frac{0}{x^2+0}=0\) and for \(x=0\) we have \(\displaystyle\lim\limits_{y\to0}\frac{0}{0+y^2}=0\).
However, for \(y=x\) we have \(\displaystyle\lim\limits_{x\to0}\frac{x^2}{x^2+x^2}=\frac{1}{2}\neq0\).

3) \(\displaystyle\lim\limits_{(x,y)\to(0,0)}\frac{x^2y}{x^4+y^2}\) does not exist. For \(y=mx\) we have \(\displaystyle\lim\limits_{x\to0}\frac{mx^3}{x^4+m^2x^2}=\lim\limits_{x\to0}\frac{mx}{x^2+m^2}=0\).
However, for \(y=x^2\) we have \(\displaystyle\lim\limits_{x\to0}\frac{x^2\cdot x^2}{x^4+x^4}=\lim\limits_{x\to0}\frac{x^4}{2x^4}=\frac{1}{2}\neq0\).

4) \(\displaystyle\lim\limits_{(x,y)\to(0,0)}\frac{2xy^2}{x^2+y^2}=0\).

Proof: Let \(\epsilon > 0\). We want to find \(\delta > 0\) such that

\[\left|\frac{2xy^2}{x^2+y^2}-0\right|<\epsilon\quad\text{for}\quad0 < \sqrt{x^2+y^2} < \delta.\]

Now we have: \(|x|=\sqrt{x^2}\leq\sqrt{x^2+y^2}\) and \(y^2\leq x^2+y^2\) and therefore

\[\left|\frac{2xy^2}{x^2+y^2}-0\right|\leq\frac{2\cdot\sqrt{x^2+y^2}\cdot(x^2+y^2)}{x^2+y^2}=2\sqrt{x^2+y^2}<2\delta\leq\epsilon\]

for \(\delta\leq\frac{1}{2}\epsilon\).

Definition: A function \(f\) of two variables is called continuous at \((a,b)\) if \(\displaystyle\lim\limits_{(x,y)\to(a,b)}f(x,y)=f(a,b)\).
We say that \(f\) is continuous on \(D\) if \(f\) is continuous at every point \((a,b)\) in \(D\).

Example: The function \(f(x,y)=\left\{\begin{array}{ll}\displaystyle\frac{2xy^2}{x^2+y^2},&(x,y)\neq(0,0)\\[2.5mm]0,&(x,y)=(0,0)\end{array}\right.\quad \) is continuous on \(\mathbb{R}^2\).

Definition: Let \(f\) be a function of three variables whose domain \(D\) includes points arbitrarily close to \((a,b,c)\). Then

\[\lim\limits_{(x,y,z)\to(a,b,c)}f(x,y,z)=L,\]

if for every \(\epsilon > 0\) there is a \(\delta > 0\) such that

\[|f(x,y,z)-L| < \epsilon\]

for all \((x,y,z)\in D\) with \(0 < \sqrt{(x-a)^2+(y-b)^2+(z-c)^2} < \delta\).

Definition: A function \(f\) of three variables is called continuous at \((a,b,c)\) if \(\displaystyle\lim\limits_{(x,y,z)\to(a,b,c)}f(x,y,z)=f(a,b,c)\).
We say that \(f\) is continuous on \(D\) if \(f\) is continuous at every point \((a,b,c)\) in \(D\).

Examples

1) \(\displaystyle\lim\limits_{(x,y,z)\to(0,0,0)}\frac{x^2+y^2}{x^2+y^2+z^2}\) does not exist, since

\[\lim\limits_{x\to0}\frac{x^2+0}{x^2+0+0}=1\quad\text{and}\quad\lim\limits_{z\to0}\frac{0+0}{0+0+z^2}=0.\]

2) \(\displaystyle\lim\limits_{(x,y,z)\to(0,0,0)}\frac{x^2y}{\sqrt{x^2+y^2+z^2}}=0\), since

\[\left|\frac{x^2y}{\sqrt{x^2+y^2+z^2}}-0\right|\leq\frac{(x^2+y^2+z^2)\sqrt{x^2+y^2+z^2}}{\sqrt{x^2+y^2+z^2}}=x^2+y^2+z^2<\delta^2\leq\epsilon.\]

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Last modified on September 15, 2021