Advanced Calculus (I)
WEN-CHING LIEN
Department of Mathematics National Cheng Kung University
WEN-CHINGLIEN Advanced Calculus (I)
3.1 Two-Sided Limits
Definition
Let a∈R, Let I be an open interval that contains a, and let f be a real function defined everywhere on I except possibly at a. Then f(x)is said to converge to L, as x approaches a, if and only if for everyǫ >0 there is aδ >0 (which in general depends on ǫ, f, I and a) such that
0< |x −a| < δ implies |f(x) −L| < ǫ.
In this case we write
L= limf(x)
3.1 Two-Sided Limits
Definition
Let a∈R, Let I be an open interval that contains a, and let f be a real function defined everywhere on I except possibly at a. Then f(x)is said to converge to L, as x approaches a, if and only if for everyǫ >0 there is aδ >0 (which in general depends on ǫ, f, I and a) such that
0< |x −a| < δ implies |f(x) −L| < ǫ.
In this case we write
L= lim
x→af(x)
and call L the limit of f(x)as x approaches a.
WEN-CHINGLIEN Advanced Calculus (I)
3.1 Two-Sided Limits
Definition
Let a∈R, Let I be an open interval that contains a, and let f be a real function defined everywhere on I except possibly at a. Then f(x)is said to converge to L, as x approaches a, if and only if for everyǫ >0 there is aδ >0 (which in general depends on ǫ, f, I and a) such that
0< |x −a| < δ implies |f(x) −L| < ǫ.
In this case we write
L= limf(x)
Example:
1. f(x) =3, lim
x→1f(x) =?
2. f(x) =3x, lim
x→1f(x) =?
3. f(x) =x2, lim
x→1f(x) =?
4. f(x) =√ x, lim
x→1f(x) =?
WEN-CHINGLIEN Advanced Calculus (I)
Example:
1. f(x) =3, lim
x→1f(x) =?
2. f(x) =3x, lim
x→1f(x) =?
3. f(x) =x2, lim
x→1f(x) =?
4. f(x) =√ x, lim
x→1f(x) =?
Example:
1. f(x) =3, lim
x→1f(x) =?
2. f(x) =3x, lim
x→1f(x) =?
3. f(x) =x2, lim
x→1f(x) =?
4. f(x) =√ x, lim
x→1f(x) =?
WEN-CHINGLIEN Advanced Calculus (I)
Example:
1. f(x) =3, lim
x→1f(x) =?
2. f(x) =3x, lim
x→1f(x) =?
3. f(x) =x2, lim
x→1f(x) =?
4. f(x) =√ x, lim
x→1f(x) =?
Example:
1. f(x) =3, lim
x→1f(x) =?
2. f(x) =3x, lim
x→1f(x) =?
3. f(x) =x2, lim
x→1f(x) =?
4. f(x) =√ x, lim
x→1f(x) =?
WEN-CHINGLIEN Advanced Calculus (I)
Remark:
Let a∈R, let I be an open interval that contains a, and let f,g be real functions defined everywhere on I except
possibly at a. If f(x) = g(x)for all x ∈I\ {a}and f(x) →L as x →a, then g(x)also has a limit as x →a, and
x→alimg(x) = lim
x→af(x).
Remark:
Let a∈R, let I be an open interval that contains a, and let f,g be real functions defined everywhere on I except
possibly at a. If f(x) = g(x)for all x ∈I\ {a}and f(x) →L as x →a, then g(x)also has a limit as x →a, and
x→alimg(x) = lim
x→af(x).
WEN-CHINGLIEN Advanced Calculus (I)
Example:
g(x) = x3+x2−x−1 x2−1 , lim
x→1g(x) =?
Example:
g(x) = x3+x2−x−1 x2−1 , lim
x→1g(x) =?
WEN-CHINGLIEN Advanced Calculus (I)
Theorem (Sequential Characterization of Limits)
Let a∈R, let I be an open interval that contains a, and let f be a real function defined everywhere on I except
possibly at a. Then
L= lim
x→af(x)
exists if and only if f(xn) →L as n→ ∞for every sequence xn ∈I\ {a}that converges to a as n→ ∞.
Theorem (Sequential Characterization of Limits)
Let a∈R, let I be an open interval that contains a, and let f be a real function defined everywhere on I except
possibly at a. Then
L= lim
x→af(x)
exists if and only if f(xn) →L as n→ ∞for every sequence xn ∈I\ {a}that converges to a as n→ ∞.
WEN-CHINGLIEN Advanced Calculus (I)
Example:
Prove that
f(x) = (
sin(1
x) x 6=0
0 x =0
has no limit as x →0.
Example:
Prove that
f(x) = (
sin(1
x) x 6=0
0 x =0
has no limit as x →0.
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2
Proof:
By examing the graph of y =f(x)(see Figure 3.1),we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2
Proof:
By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2
Proof:
By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6,the limit of f(x), as x →0, cannot exist. 2
Proof:
By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2
WEN-CHINGLIEN Advanced Calculus (I)
Proof:
By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6,the limit of f(x), as x →0, cannot exist. 2
Proof:
By examing the graph of y =f(x)(see Figure 3.1), we are led to consider two extremes:
an:= 2
(4n+1)π and bn := 2
(4n+3)π, n ∈N.
Clearly, both anand bn converge to 0 as n→ ∞. On the other hand, Since f(an) =1 and f(bn) = −1 for all n ∈N, f(an) →1 and f(bn) → −1 as n → ∞. Thus by Theorem 3.6, the limit of f(x), as x →0, cannot exist. 2
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
Suppose that a∈ R, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x
approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).
In fact,
x→alim(f +g)(x) = lim
x→af(x) + lim
x→ag(x),
xlim→a(αf)(x) = αlim
x→af(x),
x→alim(fg)(x) = lim
x→af(x)lim
x→ag(x), and (when the limit of g(x) is nonzero)
Theorem
Suppose that a∈ R, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x
approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).
In fact,
x→alim(f +g)(x) = lim
x→af(x) + lim
x→ag(x),
xlim→a(αf)(x) = αlim
x→af(x),
x→alim(fg)(x) = lim
x→af(x)lim
x→ag(x), and (when the limit of g(x) is nonzero)
x→alim
f g
(x) = limx→af(x) limx→ag(x).
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
Suppose that a∈ R, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x
approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).
In fact,
x→alim(f +g)(x) = lim
x→af(x) + lim
x→ag(x),
xlim→a(αf)(x) = αlim
x→af(x),
x→alim(fg)(x) = lim
x→af(x)lim
x→ag(x), and (when the limit of g(x) is nonzero)
Theorem
Suppose that a∈ R, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x
approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).
In fact,
x→alim(f +g)(x) = lim
x→af(x) + lim
x→ag(x),
xlim→a(αf)(x) = αlim
x→af(x),
x→alim(fg)(x) = lim
x→af(x)lim
x→ag(x), and (when the limit of g(x) is nonzero)
x→alim
f g
(x) = limx→af(x) limx→ag(x).
WEN-CHINGLIEN Advanced Calculus (I)
Theorem
Suppose that a∈ R, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f(x) and g(x) converge as x
approaches a, then so do (f +g)(x),(fg)(x),(αf)(x), and (f/g)(x)(when the limit of g(x) is nonzero).
In fact,
x→alim(f +g)(x) = lim
x→af(x) + lim
x→ag(x),
xlim→a(αf)(x) = αlim
x→af(x),
x→alim(fg)(x) = lim
x→af(x)lim
x→ag(x), and (when the limit of g(x) is nonzero)
Theorem (Squeeze Theorem For Functions) Suppose that a∈R, that I is an open interval that contains a, and that f,g,h are real functions defined everywhere on I except possibly at a.
(i)
If g(x) ≤h(x) ≤f(x)for all x ∈I\ {a}, and
x→alimf(x) = lim
x→ag(x) =L, then the limit of h(x) exists, as x →a, and
xlim→ah(x) =L.
WEN-CHINGLIEN Advanced Calculus (I)
Theorem (Squeeze Theorem For Functions) Suppose that a∈R, that I is an open interval that contains a, and that f,g,h are real functions defined everywhere on I except possibly at a.
(i)
If g(x) ≤h(x) ≤f(x)for all x ∈I\ {a}, and
x→alimf(x) = lim
x→ag(x) =L, then the limit of h(x) exists, as x →a, and
Theorem (Squeeze Theorem For Functions) Suppose that a∈R, that I is an open interval that contains a, and that f,g,h are real functions defined everywhere on I except possibly at a.
(i)
If g(x) ≤h(x) ≤f(x)for all x ∈I\ {a}, and
x→alimf(x) = lim
x→ag(x) =L, then the limit of h(x) exists, as x →a, and
xlim→ah(x) =L.
WEN-CHINGLIEN Advanced Calculus (I)
Theorem (Squeeze Theorem For Functions) Suppose that a∈R, that I is an open interval that contains a, and that f,g,h are real functions defined everywhere on I except possibly at a.
(i)
If g(x) ≤h(x) ≤f(x)for all x ∈I\ {a}, and
x→alimf(x) = lim
x→ag(x) =L, then the limit of h(x) exists, as x →a, and
Theorem (ii)
If|g(x)| ≤M for all x ∈I\ {a}and f(x) →0 as x →a, then
x→alimf(x)g(x) =0.
WEN-CHINGLIEN Advanced Calculus (I)
Theorem (ii)
If|g(x)| ≤M for all x ∈I\ {a}and f(x) →0 as x →a, then
x→alimf(x)g(x) =0.
Theorem (ii)
If|g(x)| ≤M for all x ∈I\ {a}and f(x) →0 as x →a, then
x→alimf(x)g(x) =0.
WEN-CHINGLIEN Advanced Calculus (I)
Theorem (Comparison Theorem For Functions) Suppose that a∈R, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f and g have a limit as x approaches a and
f(x) ≤g(x), x ∈I\ {a}, then
x→alimf(x) ≤ lim
x→ag(x).
Theorem (Comparison Theorem For Functions) Suppose that a∈R, that I is an open interval that contains a, and that f,g are real functions defined everywhere on I except possibly at a. If f and g have a limit as x approaches a and
f(x) ≤g(x), x ∈I\ {a}, then
x→alimf(x) ≤ lim
x→ag(x).
WEN-CHINGLIEN Advanced Calculus (I)
Example:
For each function f define the positive part of f by f+(x) = |f(x)| +f(x)
2 , x ∈Dom(f), and the negative part by
f−(x) = |f(x)| −f(x)
2 , x ∈Dom(f).
Example:
For each function f define the positive part of f by f+(x) = |f(x)| +f(x)
2 , x ∈Dom(f), and the negative part by
f−(x) = |f(x)| −f(x)
2 , x ∈Dom(f).
WEN-CHINGLIEN Advanced Calculus (I)
(a)
Prove that f+(x) ≥0, f−(x) ≥0, f(x) =f+(x) −f−(x), and
|f(x)| =f+(x) +f−(x)hold for all x ∈Dom(f).(Compare with Exercise 1,p.11.)
(b)
Prove that if
L= lim
x→af(x)
exists, then f+(x) →L+and f−(x) →L− as x →a.
(a)
Prove that f+(x) ≥0, f−(x) ≥0, f(x) =f+(x) −f−(x), and
|f(x)| =f+(x) +f−(x)hold for all x ∈Dom(f).(Compare with Exercise 1,p.11.)
(b)
Prove that if
L= lim
x→af(x)
exists, then f+(x) →L+and f−(x) →L− as x →a.
WEN-CHINGLIEN Advanced Calculus (I)
(a)
Prove that f+(x) ≥0, f−(x) ≥0, f(x) =f+(x) −f−(x), and
|f(x)| =f+(x) +f−(x)hold for all x ∈Dom(f).(Compare with Exercise 1,p.11.)
(b)
Prove that if
L= lim
x→af(x)
exists, then f+(x) →L+and f−(x) →L− as x →a.
(a)
Prove that f+(x) ≥0, f−(x) ≥0, f(x) =f+(x) −f−(x), and
|f(x)| =f+(x) +f−(x)hold for all x ∈Dom(f).(Compare with Exercise 1,p.11.)
(b)
Prove that if
L= lim
x→af(x)
exists, then f+(x) →L+and f−(x) →L− as x →a.
WEN-CHINGLIEN Advanced Calculus (I)
Example:
Let f,g be real functions, and for each
x ∈Dom(f) ∩Dom(g)define(f∨g)(x) :=max{f(x),g(x)}
and(f ∨g)(x) :=min{f(x),g(x)}.
Example:
Let f,g be real functions, and for each
x ∈Dom(f) ∩Dom(g)define(f∨g)(x) :=max{f(x),g(x)}
and(f ∨g)(x) :=min{f(x),g(x)}.
WEN-CHINGLIEN Advanced Calculus (I)
(a)
Prove that
(f ∨g)(x) = (f +g)(x) + |(f −g)(x)|
2 and
(f ∧g)(x) = (f +g)(x) − |(f −g)(x)|
2 for all x ∈Dom(f) ∩Dom(g).
(a)
Prove that
(f ∨g)(x) = (f +g)(x) + |(f −g)(x)|
2 and
(f ∧g)(x) = (f +g)(x) − |(f −g)(x)|
2 for all x ∈Dom(f) ∩Dom(g).
WEN-CHINGLIEN Advanced Calculus (I)
(b)
Prove that if
L= lim
x→af(x) and M = lim
x→ag(x)
exist, then (f ∨g)(x) →L∨M and(f ∧g)(x) →L∧M as x →a.
(b)
Prove that if
L= lim
x→af(x) and M = lim
x→ag(x)
exist, then (f ∨g)(x) →L∨M and(f ∧g)(x) →L∧M as x →a.
WEN-CHINGLIEN Advanced Calculus (I)