Limits and Continuity
Let \(f\) be a function of two variables whose domain \(D\) includes points arbitrarily close \((a,b)\). Then, we say
\[\lim\_{(x,y)\rightarrow (a,b)}f(x,y) = L\]if for every number \(\epsilon > 0\), there’s a corresponding \(\delta > 0\) such that
if \((x,y) \in D\) and \(0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta\), then \(\lvert f(x,y) - L \rvert < \epsilon\)
Existence of Limit
Unlike univariate function where we can test by \(\lim_{x \rightarrow a^{+}} f(x)= \lim_{x \rightarrow a^{-}}f(x)\) because there’re only two directions.
However, it’s not that simple for multivariate functions since there’re infinite number of directions.
Definition
If \(f(x,y) \rightarrow \rightarrow L_1\) as \((x,y) \rightarrow (a,b)\) along a path \(C_1\),
and \(f(x,y) \rightarrow \rightarrow L_2\) as \((x,y) \rightarrow (a,b)\) along a path \(C_2\),
where \(L_1 \neq L_2\), then \(\lim_{x,y \rightarrow a,b}f(x)\) does NOT exist.
Example. Show \(\lim_{x,y \rightarrow 0,0} \frac{x^2-y^2}{x^2 + y^2}\) does not exit.
- Approach along x-axis
-> \(y=0\) gives \(f(x,0)=1\), so \(\lim_{x,y \rightarrow 0,0}f(x,0) = 1\)
- Approach along y-axis
-> \(x=0\) gives \(f(0,y) = -1\) so \(\lim_{x,y \rightarrow 0,0}f(0,y)=-1\)
Since \(1 \neq -1\), the limit does not exist for \(x,y \rightarrow 0,0\).
Continuity
A function of two variables is called continuous at \((a,b)\) if
\[\lim\_{x,y \rightarrow a,b}f(x,y) = f(a,b)\]We say \(f\) is is continuous on \(D\) if \(f\) is continuous at every point \((a,b)\) in \(D\).
Exercises
Q13
Find the limit, if it exists, or show that the limit does not exit.
\[\lim\_{x,y \rightarrow 0,0} \frac{xy}{\sqrt{x^2+y^2}}\]To prove that the limit exists, we use the definition of the existence of limit.
Choose any \(\epsilon > 0\). We must show that there exists \(\delta > 0\) such that if \(0 < \sqrt{x^2 + y^2} < \delta\), then \(\rvert \frac{xy}{\sqrt{x^2+y^2}} \rvert < \epsilon\).
\[x^2 \leq x^2 + y^2\] \[\lvert x \rvert \leq \sqrt{x^2 + y^2}\] \[\frac{\lvert x \rvert}{\sqrt{x^2+y^2}} \leq 1\] \[\frac{\lvert x \rvert \lvert y \rvert}{\sqrt{x^2+y^2}} \leq \lvert y \rvert\]Also,
\[y^2 \leq x^2 + y^2\] \[\lvert y \rvert \leq \sqrt{x^2+y^2}\]Therefore,
\[\frac{\lvert x \rvert \lvert y \rvert}{\sqrt{x^2+y^2}} \leq \lvert y \rvert \leq \sqrt{x^2+y^2}\]We made an assumption: \(0 < \sqrt{x^2 + y^2} < \delta\). Hence,
\[\frac{\lvert x \rvert \lvert y \rvert}{\sqrt{x^2+y^2}} \leq \lvert y \rvert \leq \sqrt{x^2+y^2} < \delta\]If we choose \(\delta = \epsilon\). Then,
\[\frac{\lvert x \rvert \lvert y \rvert}{\sqrt{x^2+y^2}} < \delta\]We showed that the limit indeed exists.
Q17
Find the limit, if it exists, or show that the limit does not exist.
\[\lim\_{x,y \rightarrow 0,0} \frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}\]For approaching along x-axis (y=0),
\[\lim\_{x \rightarrow 0}\frac{x^2}{\sqrt{x^2+1}-1} = 0\]For approaching along \(y=\sqrt{2x}\),
\[\lim\_{x \rightarrow 0} \frac{x^2 + 2x}{x+1-1} = 2\]Since they’re different, the limit does not exist.
Q37
Q39
Use polar coordinates to find the limit.
\[\lim\_{x,y \rightarrow 0,0} \frac{x^3+y^3}{x^2+y^2}\]In polar coordinates, \(x=r\cos{\theta}, y=r\sin{\theta}\).
\[\lim\_{r \rightarrow 0^{+}}\frac{r^3\cos^3{\theta} + r^3\sin^3{\theta}}{r^2\cos^2{\theta}+r^2\sin^2{\theta}}\] \[\lim\_{r \rightarrow 0^{+}}r(cos^3{\theta}+sin^3{\theta})\] \[= 0\]Q41
Use polar coordinates to find the limit.
\[\lim\_{x,y \rightarrow 0,0} \frac{e^{-x^2-y^2}-1}{x^2+y^2}\]Using the similar approach, we can show that the answer is \(-1\).
Q45
Show that the function \(f\) giben by \(f(x) = \lvert x \rvert\) is continuous on \(\mathbb{R}^n\). Hint: Consider \(\lvert x - a \rvert^2 = (x - a) \cdot (x-a)\).
We must show the limit \(\lim_{x \rightarrow a}f(x) = L\) exists at an arbitrary \(a\).
Then, we need to show that for every \(\epsilon > 0\), there exists a corresponding \(\delta > 0\) such that if \(0 < \lvert x - a \rvert < \delta\), then \(\lvert f(x) - L \rvert < \epsilon\).
By the reverse triangle inequality,
\[\lvert x - a \rvert \geq \lvert \lvert x \rvert - \lvert a \rvert \rvert\]By our assumption \(0 < \lvert x - a \vert < \delta\), we have
\[\lvert \lvert x \rvert - \lvert a \rvert \rvert \leq \lvert x - a \rvert < \delta\]Since \(f(x) = \lvert x \rvert\) and \(\lim_{x \rightarrow a}\lvert x \rvert = \lvert a \rvert = L\)
\[\lvert f(x) - L \rvert < \delta\]Hence, \(f(x)\) is continuous on \(\mathbb{R}^n\).
References
[1] Stewart Calculus 8th Edition