True False
We investigate what continuity means for functions of several variables.
Now that we have defined limits, we can define continuity.
Let \(F:\R ^n\to \R \) and \(\pt {a}\) be an interior point of the domain of \(F\). We say \(F\) is continuous at \(\pt {x}=\pt {a}\),
if
- \(F(\pt {a})\) exists.
- \(\lim _{\pt {x}\to \pt {a}} F(\pt {x})\) exists.
- \(\lim _{\pt {x}\to \pt {a}} F(\pt {x}) = F(\pt {a}).\)
\(F\) is continuous on an open set \(S\) if \(F\) is continuous at all \(\pt {x} \in S\).
The limit laws can be used to write corresponding continuity laws.
Limit Laws Let \(F:\R ^n\to \R \) and \(G:\R ^n\to \R \) be continuous functions of several variables, and \(b\) be a real
number.
\begin{align*} \pt {x} &= \point {x_1,x_2,\dots ,x_n}\\ \pt {a} &= \point {a_1,a_2,\dots ,a_n}, \end{align*}
where
\[ \lim _{\pt {x}\to \pt {a}}F(\pt {x}) = L \quad \text {and}\quad \lim _{\pt {x}\to \pt {a}} G(\pt {x}) = M. \]
- Constant Law
-
\(F(\pt {x}) = b\) is continuous.
- Identity Law
-
\(F(\pt {x}) = x_i\) is continuous.
- Sum/Difference Law
-
\(F(\pt {x})\pm G(\pt {x})\) is continuous.
- Scalar Multiple Law
-
\(b\cdot F(\pt {x})\) is continuous.
- Product Law
-
\(F(\pt {x})\cdot G(\pt {x})\) is continuous.
- Quotient Law
-
\(\frac {F(\pt {x})}{G(\pt {x})}\) is continuous where \(G(\pt {x}) \neq 0\).
True or false: If \(F:\R ^2\to \R \) and \(G:\R ^2\to \R \) are continuous functions on an open disk \(B\), then \(F\pm G\) is
continuous on \(B\).
True or false: If \(F:\R ^2\to \R \) and \(G:\R ^2\to \R \) are continuous functions on an open disk \(B\), then \(F/G\) is
continuous on \(B\).
True False
The function \(F/G\) may or may not be continuous, it
depends on whether \(G(x,y)=0\). If \(G(x,y)=0\), then \(F/G\) not continuous at that point.
Composition Limit Law Let \(f:\R \to \R \) be a continuous function on an interval \(I\). Let \(G:\R ^n\to \R \) be a
function whose range is contained in (or equal to) \(I\), Then
\[ \lim _{\pt {x}\to \pt {a}} f( G(\pt {x})) = f(\lim _{\pt {x}\to \pt {a}}G(\pt {x})) \]
Composition of Composite Functions Let \(G:\R ^n\to \R \) be continuous on an open disk \(B\), where the
range of \(G\) on \(B\) is \(I\), and let \(f\) be a single variable function that is continuous on \(I\). Then
\[ f\circ G(\pt {x}) =f(G(\pt {x})), \]
is
continuous on \(B\). Show that the function
\[ F(x,y) = \sin (x^2\cos (y)) \]
is continuous for all points in \(\R ^2\). Let
\[ F_1(x,y) = x^2. \]
Since \(y\) is not actually
used in the function, and polynomials are continuousare not continuous, we
conclude \(F_1\) is continuous everywhere. A similar statement can be made about \[ F_2(x,y) = \cos (y). \]
Setting \[ F_3=F_1\cdot F_2 \]
we obtain a continuous function from \(\R ^2\to \R \). Since sine is continuousis not
continuous for all real values, the composition of sine with \(F_3\) is continuous.
Hence, \(\sin (F_3(x,y)) = \sin (x^2\cos y)\) is continuous everywhere. We finish by presenting you with a plot of
\(F\):
Let
\[ F(x,y) = \begin{cases} \frac {\cos (y)\sin (x)}{x} & x\neq 0 \\ \cos (y) & x=0 \end{cases} \]
Is \(F\) continuous at \((0,0)\)? Is \(F\) continuous everywhere? To determine if \(F\) is continuous at \((0,0)\),
we need to compare
\[ \lim _{\point {x,y}\to \point {0,0}} F(x,y)\quad \text {to}\quad F(0,0). \]
Applying the definition of \(F\), we see that: \[ F(0,0) = \answer [given]{1} \]
We now consider the
limit \[ \lim _{\point {x,y}\to \point {0,0}}F(x,y). \]
Substituting \(0\) for \(x\) and \(y\) in \((\cos (y)\sin (x))/x\) returns the indeterminate form \(\relax \boldsymbol {\tfrac {0}{0}}\), so we need to do
more work to evaluate this limit.
Consider two related limits:
\begin{align*} \lim _{\point {x,y}\to \point {0,0}} \cos (y)\\ \lim _{\point {x,y}\to \point {0,0}} \frac {\sin (x)}{x}. \end{align*}
The first limit does not contain \(x\), and since \(\cos (y)\) is continuous,
\begin{align*} \lim _{\point {x,y}\to \point {0,0}} \cos (y) &=\lim _{y\to 0} \cos (y) \\ &=\answer [given]{1} \end{align*}
The second limit does not contain \(y\). But we know
\begin{align*} \lim _{\point {x,y}\to \point {0,0}} \frac {\sin (x)}{x} &= \lim _{x\to 0} \frac {\sin (x)}{x} \\ &= \answer [given]{1}. \end{align*}
Finally, we know that we can combine these two limits so that \(\lim _{\point {x,y}\to \point {0,0}} \frac {\cos (y)\sin (x)}{x}\)
\begin{align*} &= \lim _{\point {x,y}\to \point {0,0}} (\cos (y))\left (\frac {\sin (x)}{x}\right ) \\ &=\left (\lim _{\point {x,y}\to \point {0,0}} \cos (y)\right )\left (\lim _{\point {x,y}\to \point {0,0}} \frac {\sin (x)}{x}\right ) \\ &=\answer [given]{1}\cdot \answer [given]{1}. \end{align*}
We have found that \(\lim _{\point {x,y}\to \point {0,0}} \frac {\cos (y)\sin (x)}{x} = F(0,0)\), so \(F\) is continuous at \((0,0)\).
A similar analysis shows that \(F\) is continuous at all points in \(\mathbb {R}^2\). As long as \(x\neq 0\), we can evaluate the limit directly; when \(x=0\), a similar analysis shows that the limit is \(\cos y\). Thus we can say that \(F\) is continuous everywhere. We finish by presenting you with a plot of \(F\):