4.1.E: Problems on Limits and Continuity
- Page ID
- 22630
Prove Corollary \(2 .\) Why can one interchange \(G_{p}(\delta)\) and \(G_{\neg p}(\delta)\) here?
Prove Corollary \(3 .\) By induction, extend its first clause to unions of \(n\) paths. Disprove it for infinite unions of paths (see Problem 9 in §3).
Prove that a function \(f : E^{1} \rightarrow\left(T, \rho^{\prime}\right)\) is continuous at \(p\) iff
\[
f(p)=f\left(p^{-}\right)=f\left(p^{+}\right) .
\]
Show that relative limits and continuity at \(p\) (over \(B )\) are equivalent to the ordinary ones if \(B\) is a neighborhood of \(p\) (Chapter 3, §12); for example, if it is some \(G_{p}\).
Discuss Figures \(13-15\) in detail, comparing \(f(p), f\left(p^{-}\right),\) and \(f\left(p^{+}\right) ;\) see Problem \(2^{\prime} .\)
Observe that in Figure \(13,\) different values of \(\delta\) result at \(p\) and \(p_{1}\) for the same \(\varepsilon .\) Thus \(\delta\) depends on both \(\varepsilon\) and the choice of \(p .\)
Complete the missing details in Examples \((\mathrm{d})-(\mathrm{g}) . \operatorname{In}(\mathrm{d}),\) redefine \(f(x)\) to be the least integer \(\geq x .\) Show that \(f\) is then left-continuous on \(E^{1}\).
Give explicit definitions (such as \((3) )\) for
\[
\begin{array}{ll}{\text { (a) } \lim _{x \rightarrow+\infty} f(x)=-\infty ;} & {\text { (b) } \lim _{x \rightarrow-\infty} f(x)=q} ; \\ {\text { (c) } \lim _{x \rightarrow p} f(x)=+\infty ;} & {\text { (d) } \lim _{x \rightarrow p} f(x)=-\infty} ; \\ {\text { (e) } \lim _{x \rightarrow p^{-}} f(x)=+\infty ;} & {\text { (f) } \lim _{x \rightarrow p^{+}} f(x)=-\infty} . \end{array}
\]
In each case, draw a diagram (such as Figures \(13-15 )\) and determine whether the domain and range of \(f\) must both be in \(E^{*}\).
Define \(f : E^{1} \rightarrow E^{1}\) by
\[
f(x)=\frac{x^{2}-1}{x-1} \text { if } x \neq 1, \text { and } f(1)=0 .
\]
Show that \(\lim _{x \rightarrow 1} f(x)=2\) exists, yet \(f\) is discontinuous at \(p=1 .\) Make it continuous by redefining \(f(1) .\)
\([\text { Hint: For } x \neq 1, f(x)=x+1 . \text { Proceed as in Example (b), using the deleted globe }\) \(G_{\neg p}(\delta) . ]\)
Find \(\lim _{x \rightarrow p} f(x)\) and check continuity at \(p\) in the following cases, assuming that \(D_{f}=A\) is the set of all \(x \in E^{1}\) for which the given expression for \(f(x)\) has sense. Specify that set.
\[
\begin{array}{l}{\text { (a) } \lim _{x \rightarrow 2}\left(2 x^{2}-3 x-5\right) ; \quad \text { (b) } \lim _{x \rightarrow 1} \frac{3 x+2}{2 x-1}} \\ {\text { (c) } \lim _{x \rightarrow-1}\left(\frac{x^{2}-4}{x+2}-1\right) ;} & {\text { (d) } \lim _{x \rightarrow 2} \frac{x^{3}-8}{x-2}} \\ {\text { (e) } \lim _{x \rightarrow a} \frac{x^{4}-a^{4}}{x-a} ;} & {\text { (f) } \lim _{x \rightarrow 0}\left(\frac{x}{x+1}\right)^{3}} \\ {\text { (g) } \lim _{x \rightarrow-1}\left(\frac{1}{x^{2}+1}\right)^{2}}\end{array}
\]
\( [ \text { Example solution: Find } \lim _{x \rightarrow 1} \frac{5 x^{2}-1}{2 x+3} .\)
Here
\[
f(x)=\frac{5 x^{2}-1}{2 x+3} ; A=E^{1}-\left\{-\frac{3}{2}\right\} ; p=1 .
\]
We show that \(f\) is continuous at \(p,\) and so (by Corollary 2\()\)
\[
f(x) = \frac{5x^{2} - 1}{2x + 3} ; A = E^{1} - \large\{ - \frac{3}{2} \large\} ; p = 1 .
\]
We show that \(f\) is continuous at \(p\), and so (by Corollary 2)
\[
\lim _{x \rightarrow p} f(x)=f(p)=f(1)=\frac{4}{5} .
\]
Using formula \((1),\) we fix an arbitrary \(\varepsilon>0\) and look for a \(\delta\) such that
\[
\left(\forall x \in A \cap G_{p}(\delta)\right) \quad \rho(f(x), f(1))=|f(x)-f(1)|<\varepsilon, \text { i.e., },\left|\frac{5 x^{2}-1}{2 x+3}-\frac{4}{5}\right|<\varepsilon ;
\]
or, by putting everything over a common denominator and using properties of absolute values,
\[
|x-1| \frac{|25 x+17|}{5|2 x+3|}<\varepsilon \text { whenever }|x-1|<\delta \text { and } x \in A .
\]
(Usually in such problems, it is desirable to factor out \(x-p . )\)
By Note \(4,\) we may assume \(0<\delta \leq 1 .\) Then \(|x-1|<\delta\) implies \(-1 \leq x-1 \leq 1\) i.e., \(0 \leq x \leq 2,\) so
\[
5|2 x+3| \geq 15 \text { and }|25 x+17| \leq 67 .
\]
Hence (6) will certainly hold if
\[
|x-1| \frac{67}{15}<\varepsilon, \text { i.e., if }|x-1|<\frac{15 \varepsilon}{67} .
\]
To achieve it, we choose \(\delta=\min (1,15 \varepsilon / 67) .\) Then, reversing all steps, we obtain \((6),\) and hence \(\lim _{x \rightarrow 1} f(x)=f(1)=4 / 5 . ]\)
Find (using definitions, such as \((3) )\)
\[
\begin{array}{ll}{\text { (a) } \lim _{x \rightarrow+\infty} \frac{1}{x} ;} & {\text { (b) } \lim _{x \rightarrow-\infty} \frac{3 x+2}{2 x-1}} ; \\ {\text { (c) } \lim _{x \rightarrow+\infty} \frac{x^{3}}{1-x^{2}} ;} & {\text { (d) } \lim _{x \rightarrow 3^{+}} \frac{x-1}{x-3}} ; \\ {\text { (e) } \lim _{x \rightarrow 3^{-}} \frac{x-1}{x-3} ;} & {\text { (f) } \lim _{x \rightarrow 3}\left|\frac{x-1}{x-3}\right|} . \end{array}
\]
Prove that if
\[
\lim _{x \rightarrow p} f(x)=\overline{q} \in E^{n}\left(^{*} C^{n}\right) ,
\]
then for each scalar \(c\),
\[
\lim _{x \rightarrow p} c f(x)=c \overline{q} .
\]
Define \(f : E^{1} \rightarrow E^{1}\) by
\[
f(x)=x \cdot \sin \frac{1}{x} \text { if } x \neq 0, \text { and } f(0)=0 .
\]
Show that \(f\) is continuous at \(p=0,\) i.e.,
\[
\lim _{x \rightarrow 0} f(x)=f(0)=0 .
\]
Draw an approximate graph (it is contained between the lines \(y=\pm x )\).
\(\left[\text { Hint: }\left|x \cdot \sin \frac{1}{x}-0\right| \leq|x| .\right]\)
Discuss the statement: \(f\) is continuous at \(p\) iff
\[
\left(\forall G_{f(p)}\right)\left(\exists G_{p}\right) \quad f\left[G_{p}\right] \subseteq G_{f(p)} .
\]
Define \(f : E^{1} \rightarrow E^{1}\) by
\[
f(x)=x \text { if } x \text { is rational }
\]
and
\[
f(x)=0 \text { otherwise } .
\]
Show that \(f\) is continuous at 0 but nowhere else. How about relative continuity?
Let \(A=(0,+\infty) \subset E^{1} .\) Define \(f : A \rightarrow E^{1}\) by
\[
f(x)=0 \text { if } x \text { is irrational }
\]
and
\[
f(x)=\frac{1}{n} \text { if } x=\frac{m}{n}(\text {in lowest terms })
\]
for some natural \(m\) and \(n .\) Show that \(f\) is continuous at each irrational, but at no rational, point \(p \in A .\)
[Hints: If \(p\) is irrational, fix \(\varepsilon>0\) and an integer \(k>1 / \varepsilon .\) In \(G_{p}(1),\) there are only finitely many irreducible fractions
\[
\frac{m}{n}>0 \text { with } n \leq k ,
\]
so one of them, call it \(r,\) is closest to \(p .\) Put
\[
\delta=\min (1,|r-p|)
\]
and show that
\[
\left(\forall x \in A \cap G_{p}(\delta)\right) \quad|f(x)-f(p)|=f(x)<\varepsilon ,
\]
distinguishing the cases where \(x\) is rational and irrational.
If \(p\) is rational, use the fact that each \(G_{p}(\delta)\) contains irrationals \(x\) at which
\[
f(x)=0 \Longrightarrow|f(x)-f(p)|=f(p) .
\]
Take \(\varepsilon<f(p) . ]\)
Given two reals, \(p>0\) and \(q>0,\) define \(f : E^{1} \rightarrow E^{1}\) by
\[
f(0)=0 \text { and } f(x)=\left(\frac{x}{p}\right) \cdot\left[\frac{q}{x}\right] \text { if } x \neq 0 .
\]
here \([q / x]\) is the integral part of \(q / x\).
(i) Is \(f\) left or right continuous at 0\(?\)
(ii) Same question with \(f(x)=[x / p](q / x)\).
Prove that if \((S, \rho)\) is discrete, then all functions \(f : S \rightarrow\left(T, \rho^{\prime}\right)\) are continuous. What if \(\left(T, \rho^{\prime}\right)\) is discrete but \((S, \rho)\) is not?