Exercises Show that a
linear functionMDLD/linear function
R
⟶
R
,
x
⟼
a
x
,
{\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto ax,}
is continuous.
Let
D
⊆
R
{\displaystyle {}D\subseteq \mathbb {R} }
f
:
D
→
R
{\displaystyle {}f\colon D\rightarrow \mathbb {R} }
a
∈
D
{\displaystyle {}a\in D}
f
{\displaystyle {}f}
continuous MDLD/continuous (R)
in
a
{\displaystyle {}a}
For every
n
∈
N
+
{\displaystyle {}n\in \mathbb {N} _{+}}
m
∈
N
+
{\displaystyle {}m\in \mathbb {N} _{+}}
|
x
−
a
|
≤
1
m
{\displaystyle {}\vert {x-a}\vert \leq {\frac {1}{m}}\,}
the estimate
|
f
(
x
)
−
f
(
a
)
|
≤
1
n
{\displaystyle {}\vert {f(x)-f(a)}\vert \leq {\frac {1}{n}}\,}
holds.
For every
s
∈
N
{\displaystyle {}s\in \mathbb {N} }
r
∈
N
{\displaystyle {}r\in \mathbb {N} }
|
x
−
a
|
≤
1
10
r
{\displaystyle {}\vert {x-a}\vert \leq {\frac {1}{10^{r}}}\,}
the estimate
|
f
(
x
)
−
f
(
a
)
|
≤
1
10
s
{\displaystyle {}\vert {f(x)-f(a)}\vert \leq {\frac {1}{10^{s}}}\,}
holds.
Prove that the function
R
⟶
R
,
x
⟼
|
x
|
,
{\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \vert {x}\vert ,}
is continuous.
Exercise Exercise 10.4
change ===
Prove that the function
R
≥
0
⟶
R
≥
0
,
x
⟼
x
,
{\displaystyle \mathbb {R} _{\geq 0}\longrightarrow \mathbb {R} _{\geq 0},x\longmapsto {\sqrt {x}},}
is continuous.
Farmer Ernst wants to make a square-shaped field for melons. The field shall be
100
{\displaystyle {}100}
99
{\displaystyle {}99}
101
{\displaystyle {}101}
Let
f
(
x
)
=
2
x
3
−
4
x
+
5
.
{\displaystyle {}f(x)=2x^{3}-4x+5\,.}
Show that for all
x
∈
R
{\displaystyle {}x\in \mathbb {R} }
|
x
−
3
|
≤
1
800
,
{\displaystyle {}\vert {x-3}\vert \leq {\frac {1}{800}}\,,}
then
|
f
(
x
)
−
f
(
3
)
|
≤
1
10
.
{\displaystyle {}\vert {f(x)-f(3)}\vert \leq {\frac {1}{10}}\,.}
For the function
f
(
x
)
=
2
x
3
−
4
x
2
+
x
−
6
{\displaystyle {}f(x)=2x^{3}-4x^{2}+x-6\,}
and the point
a
=
1
{\displaystyle {}a=1}
ϵ
=
1
10
{\displaystyle {}\epsilon ={\frac {1}{10}}}
δ
>
0
{\displaystyle {}\delta >0}
d
(
x
,
a
)
≤
δ
{\displaystyle {}d(x,a)\leq \delta \,}
implies the estimate
d
(
f
(
x
)
,
f
(
a
)
)
≤
ϵ
.
{\displaystyle {}d(f(x),f(a))\leq \epsilon \,.}
Let
T
⊆
R
{\displaystyle {}T\subseteq \mathbb {R} }
f
:
T
⟶
R
{\displaystyle f\colon T\longrightarrow \mathbb {R} }
be a continuous function. Let
x
∈
T
{\displaystyle {}x\in T}
f
(
x
)
>
0
{\displaystyle {}f(x)>0}
f
(
y
)
>
0
{\displaystyle {}f(y)>0}
y
{\displaystyle {}y}
]
x
−
a
,
x
+
a
[
{\displaystyle {}]x-a,x+a[}
Let
a
∈
R
{\displaystyle {}a\in \mathbb {R} }
f
,
g
:
R
⟶
R
{\displaystyle f,g\colon \mathbb {R} \longrightarrow \mathbb {R} }
be continuous functions with
f
(
a
)
>
g
(
a
)
.
{\displaystyle {}f(a)>g(a)\,.}
Show that there exists some
δ
>
0
{\displaystyle {}\delta >0}
f
(
x
)
>
g
(
x
)
{\displaystyle {}f(x)>g(x)\,}
holds for all
x
∈
[
a
−
δ
,
a
+
δ
]
{\displaystyle {}x\in [a-\delta ,a+\delta ]}
Let
f
:
R
→
R
{\displaystyle {}f\colon \mathbb {R} \rightarrow \mathbb {R} }
continuous function .MDLD/continuous function (R)
Show the following statements.
The function
f
{\displaystyle {}f}
Q
{\displaystyle {}\mathbb {Q} }
The value
f
(
a
)
{\displaystyle {}f(a)}
f
(
x
)
{\displaystyle {}f(x)}
x
≠
a
{\displaystyle {}x\neq a}
If for all
x
<
a
{\displaystyle {}x<a}
f
(
x
)
≤
c
{\displaystyle {}f(x)\leq c\,}
holds, then also
f
(
a
)
≤
c
{\displaystyle {}f(a)\leq c\,}
holds.
Let
a
<
b
<
c
{\displaystyle {}a<b<c}
f
:
[
a
,
b
]
⟶
R
{\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }
and
g
:
[
b
,
c
]
⟶
R
{\displaystyle g\colon [b,c]\longrightarrow \mathbb {R} }
be continuous functions such that
f
(
b
)
=
g
(
b
)
{\displaystyle {}f(b)=g(b)}
h
:
[
a
,
c
]
⟶
R
{\displaystyle h\colon [a,c]\longrightarrow \mathbb {R} }
such that
h
(
t
)
=
f
(
t
)
for
t
≤
b
and
h
(
t
)
=
g
(
t
)
for
t
>
b
{\displaystyle h(t)=f(t){\text{ for }}t\leq b{\text{ and }}h(t)=g(t){\text{ for }}t>b}
is also continuous.
Let
f
:
[
a
,
b
]
⟶
R
{\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }
be a continuous function. Show that there exists a continuous extension
f
~
:
R
⟶
R
{\displaystyle {\tilde {f}}\colon \mathbb {R} \longrightarrow \mathbb {R} }
of
f
{\displaystyle {}f}
Let
T
⊆
R
{\displaystyle {}T\subseteq \mathbb {R} }
f
:
T
⟶
R
{\displaystyle f\colon T\longrightarrow \mathbb {R} }
be a
function .MDLD/function
Show that
f
{\displaystyle {}f}
continuous .MDLD/continuous (R)
Show that there exists a
continuous function MDLD/continuous function (R)
f
:
R
⟶
R
{\displaystyle f\colon \mathbb {R} \longrightarrow \mathbb {R} }
such that
f
{\displaystyle {}f}
[
0
,
δ
]
{\displaystyle {}[0,\delta ]}
δ
>
0
{\displaystyle {}\delta >0}
Is it possible to draw such a function? See also
Exercise 16.25
.
Compute the limit of the sequence
x
n
=
5
(
2
n
+
1
n
)
3
−
4
(
2
n
+
1
n
)
2
+
2
(
2
n
+
1
n
)
−
3
{\displaystyle {}x_{n}=5\left({\frac {2n+1}{n}}\right)^{3}-4\left({\frac {2n+1}{n}}\right)^{2}+2\left({\frac {2n+1}{n}}\right)-3\,}
for
n
→
∞
{\displaystyle {}n\rightarrow \infty }
Prove that the function
f
:
R
⟶
R
{\displaystyle f\colon \mathbb {R} \longrightarrow \mathbb {R} }
defined by
f
(
x
)
=
{
x
,
if
x
∈
Q
,
0
,
otherwise
,
{\displaystyle {}f(x)={\begin{cases}x,\,{\text{ if }}x\in \mathbb {Q} \,,\\0,\,{\text{ otherwise}}\,,\end{cases}}\,}
is only at the zero point
0
{\displaystyle {}0}
Determine the limit of the sequence
x
n
=
7
n
2
−
4
3
n
2
−
5
n
+
2
,
n
∈
N
.
{\displaystyle x_{n}={\sqrt {\frac {7n^{2}-4}{3n^{2}-5n+2}}},\,n\in \mathbb {N} .}
The sequence
(
x
n
)
n
∈
N
{\displaystyle {}{\left(x_{n}\right)}_{n\in \mathbb {N} }}
x
0
=
1
{\displaystyle {}x_{0}=1}
x
n
+
1
=
x
n
+
1
.
{\displaystyle {}x_{n+1}={\sqrt {x_{n}+1}}\,.}
Show that this sequence converges and determine its limit.
Prove directly the computing rules from
Lemma 10.6
(without referring to the sequence crtierion).
Show that the function
f
:
R
⟶
R
,
x
⟼
2
x
7
−
3
x
|
6
x
3
−
11
|
|
3
x
−
5
|
+
|
4
x
3
−
5
x
+
1
|
,
{\displaystyle f\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto {\frac {2x^{7}-3x\vert {6x^{3}-11}\vert }{\vert {3x-5}\vert +\vert {4x^{3}-5x+1}\vert }},}
is
continuous .MDLD/continuous (R)
Give an example for a
continuous function MDLD/continuous function (R)
f
:
R
≥
0
→
R
≥
0
{\displaystyle {}f\colon \mathbb {R} _{\geq 0}\rightarrow \mathbb {R} _{\geq 0}}
absolutely convergent MDLD/absolutely convergent (R)
real seriesMDLD/real series
∑
k
=
0
∞
a
k
{\displaystyle {}\sum _{k=0}^{\infty }a_{k}}
a
k
≥
0
{\displaystyle {}a_{k}\geq 0}
∑
k
=
0
∞
f
(
a
k
)
{\displaystyle {}\sum _{k=0}^{\infty }f(a_{k})}
Let
a
∈
R
{\displaystyle {}a\in \mathbb {R} }
f
,
g
,
h
:
R
→
R
{\displaystyle {}f,g,h\colon \mathbb {R} \rightarrow \mathbb {R} }
g
{\displaystyle {}g}
h
{\displaystyle {}h}
continuous MDLD/continuous (R)
in
a
{\displaystyle {}a}
g
(
a
)
=
f
(
a
)
=
h
(
a
)
{\displaystyle {}g(a)=f(a)=h(a)}
g
(
x
)
≤
f
(
x
)
≤
h
(
x
)
{\displaystyle {}g(x)\leq f(x)\leq h(x)}
x
∈
R
{\displaystyle {}x\in \mathbb {R} }
f
{\displaystyle {}f}
a
{\displaystyle {}a}
Determine the
limitMDLD/limit (Function R)
of the
rational functionMDLD/rational function (R)
x
−
1
x
2
−
1
{\displaystyle {\frac {x-1}{x^{2}-1}}}
in the point
a
=
1
{\displaystyle {}a=1}
Exercise Exercise 10.24
change ===
Let
T
⊆
R
{\displaystyle {}T\subseteq \mathbb {R} }
a
∈
R
{\displaystyle {}a\in \mathbb {R} }
f
:
T
⟶
R
{\displaystyle f\colon T\longrightarrow \mathbb {R} }
be a
function MDLD/function
and
b
∈
R
{\displaystyle {}b\in \mathbb {R} }
We have
lim
x
→
a
f
(
x
)
=
b
.
{\displaystyle {}\operatorname {lim} _{x\rightarrow a}\,f(x)=b\,.}
For every sequence
(
x
n
)
n
∈
N
{\displaystyle {}{\left(x_{n}\right)}_{n\in \mathbb {N} }}
T
{\displaystyle {}T}
converges MDLD/converges (R)
to
a
{\displaystyle {}a}
(
f
(
x
n
)
)
n
∈
N
{\displaystyle {}{\left(f(x_{n})\right)}_{n\in \mathbb {N} }}
b
{\displaystyle {}b}
Hint: This is proved similarly to the
sequence criterion for continuity .
Let
T
=
{
1
n
∣
n
∈
N
+
}
⊆
R
{\displaystyle {}T={\left\{{\frac {1}{n}}\mid n\in \mathbb {N} _{+}\right\}}\subseteq \mathbb {R} \,}
be the set of the unit fractions and let
(
x
n
)
n
∈
N
{\displaystyle {}{\left(x_{n}\right)}_{n\in \mathbb {N} }}
b
∈
R
{\displaystyle {}b\in \mathbb {R} }
D
=
T
∪
{
0
}
{\displaystyle {}D=T\cup \{0\}}
The sequence
converges MDLD/converges (R)
to
b
{\displaystyle {}b}
The function
f
:
T
⟶
R
{\displaystyle f\colon T\longrightarrow \mathbb {R} }
given by
f
(
1
n
)
=
x
n
{\displaystyle {}f{\left({\frac {1}{n}}\right)}=x_{n}\,}
has a
limit MDLD/limit (real function)
lim
x
→
0
f
(
x
)
=
b
{\displaystyle {}\operatorname {lim} _{x\rightarrow 0}\,f(x)=b}
The function
f
~
:
D
⟶
R
{\displaystyle {\tilde {f}}\colon D\longrightarrow \mathbb {R} }
given by
f
~
(
1
n
)
=
x
n
{\displaystyle {}{\tilde {f}}{\left({\frac {1}{n}}\right)}=x_{n}\,}
and
f
~
(
0
)
=
b
{\displaystyle {}{\tilde {f}}(0)=b}
Hand-in-exercises For the function
f
(
x
)
=
x
3
+
5
x
2
−
3
x
+
2
{\displaystyle {}f(x)=x^{3}+5x^{2}-3x+2\,}
and the point
a
=
3
{\displaystyle {}a=3}
ϵ
=
1
100
{\displaystyle {}\epsilon ={\frac {1}{100}}}
δ
>
0
{\displaystyle {}\delta >0}
d
(
x
,
a
)
≤
δ
{\displaystyle {}d(x,a)\leq \delta \,}
implies the estimate
d
(
f
(
x
)
,
f
(
a
)
)
≤
ϵ
.
{\displaystyle {}d(f(x),f(a))\leq \epsilon \,.}
We consider the function
f
(
x
)
=
{
1
for
x
≤
−
1
,
x
2
for
−
1
<
x
<
2
,
−
2
x
+
7
for
x
≥
2
.
{\displaystyle {}f(x)={\begin{cases}1{\text{ for }}x\leq -1\,,\\x^{2}{\text{ for }}-1<x<2\,,\\-2x+7{\text{ for }}x\geq 2\,.\end{cases}}\,}
Determine the points
x
∈
R
{\displaystyle {}x\in \mathbb {R} }
f
{\displaystyle {}f}
continuous .MDLD/continuous (R)
Prove that the function
f
:
R
→
R
{\displaystyle {}f\colon \mathbb {R} \rightarrow \mathbb {R} }
f
(
x
)
=
{
1
,
if
x
∈
Q
,
0
otherwise
,
{\displaystyle {}f(x)={\begin{cases}1,{\text{ if }}x\in \mathbb {Q} \,,\\0\,{\text{ otherwise}}\,,\end{cases}}\,}
is for no point
x
∈
R
{\displaystyle {}x\in \mathbb {R} }
Compute the limit of the sequence
b
n
=
2
a
n
4
−
6
a
n
3
+
a
n
2
−
5
a
n
+
3
,
{\displaystyle {}b_{n}=2a_{n}^{4}-6a_{n}^{3}+a_{n}^{2}-5a_{n}+3\,,}
where
a
n
=
3
n
3
−
5
n
2
+
7
4
n
3
+
2
n
−
1
.
{\displaystyle {}a_{n}={\frac {3n^{3}-5n^{2}+7}{4n^{3}+2n-1}}\,.}
Determine the
limitMDLD/limit (Function R)
of the
rational functionMDLD/rational function (R)
2
x
3
+
3
x
2
−
1
x
3
−
x
2
+
x
+
3
{\displaystyle {\frac {2x^{3}+3x^{2}-1}{x^{3}-x^{2}+x+3}}}
in the point
a
=
−
1
{\displaystyle {}a=-1}
![{\displaystyle {}]x-a,x+a[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6dd61fb109cd49187d5a310f1df93c2ddb976cb)
![{\displaystyle {}x\in [a-\delta ,a+\delta ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c9b3bafaa2faac72bc38dad249dc53053ba0cb2)
![{\displaystyle f\colon [a,b]\longrightarrow \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bda17a8ca7ea71906b37514dd0342131a206a98)
![{\displaystyle g\colon [b,c]\longrightarrow \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba3a0a85a2d69af0e078a54574e7f9fda0dbebe6)
![{\displaystyle h\colon [a,c]\longrightarrow \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/73001510e89247c7bfa30dc505633d98e0d8b5be)
![{\displaystyle {}[0,\delta ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9890e90b6d9fed88581fe88878da21efee2105dd)
