THEOREM 3.2 that f is a function that is defined on I except possibly at the point c. hat f is a function that is defined on I except possibly at the point c. Let I be an open interval that contains the point c and suppose a) The function f has limit L at c if and only if for each sequence {xn} in I\{c} that converges to c, the sequence (f (x)} converges to L. b) Suppose that there are sequences (x} and (y} in I\c} that converge to c such that {f (xn)} converges to L, and (f(yn)} converges to L2. If Li + L2, then the functionf does not have a limit at c. Proof. Let P be the statement "the function f has limit L at c" and let Q be the statement “for each sequence (x,} in I\ {c} that converges to c, the sequence If (Xn)} converges to L". A proof that P = Q will be left as an exercise. To that Q = P, we will prove the contrapositive. Suppose lim f (x) # L. By the negation of the definition, there exists e >0 prove such that for each 8 > 0 there is a point x EI such that 0 < |x – c| < 8 and f(x) – L| E. n particular, for each positive integer n, there exists a point x, E I such that 0 < Įxn – c| < 1/n and |f(x,) – L 2 E. ow {Xn} is a sequence in I \ {c} that converges to c, but the sequence (f(x)} does ot converge to L. This completes the proof of part (a). Part (b) of the theorem
THEOREM 3.2 that f is a function that is defined on I except possibly at the point c. hat f is a function that is defined on I except possibly at the point c. Let I be an open interval that contains the point c and suppose a) The function f has limit L at c if and only if for each sequence {xn} in I\{c} that converges to c, the sequence (f (x)} converges to L. b) Suppose that there are sequences (x} and (y} in I\c} that converge to c such that {f (xn)} converges to L, and (f(yn)} converges to L2. If Li + L2, then the functionf does not have a limit at c. Proof. Let P be the statement "the function f has limit L at c" and let Q be the statement “for each sequence (x,} in I\ {c} that converges to c, the sequence If (Xn)} converges to L". A proof that P = Q will be left as an exercise. To that Q = P, we will prove the contrapositive. Suppose lim f (x) # L. By the negation of the definition, there exists e >0 prove such that for each 8 > 0 there is a point x EI such that 0 < |x – c| < 8 and f(x) – L| E. n particular, for each positive integer n, there exists a point x, E I such that 0 < Įxn – c| < 1/n and |f(x,) – L 2 E. ow {Xn} is a sequence in I \ {c} that converges to c, but the sequence (f(x)} does ot converge to L. This completes the proof of part (a). Part (b) of the theorem
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![mathematically before reading the proof. (It might be helpful to read the discussion
that follows Theorem 2.4 or to look at Appendix A.)
THEOREM 3.2
THEIS a function that is defined on I except possibly at the point c.
Let I be an open interval that contains the point c and suppose
a) The function f has limit L at c if and only if for each sequence {Xn) in
I\ {c} that converges to c, the sequence (f (x,)} converges to L.
b) Suppose that there are sequences (x,} and (y,} in I\ {c} that converge
to c such that {ƒ(xn)} converges to L, and (f(yn)} converges to L2. If
Li + L2, then the functionf does not have a limit at c.
Proof Let P be the statement "the function f has limit L at c" and let Q be
the statement "for each sequence {xn} in I\ {c} that converges to c, the sequence
(f(xn)} converges to L". A proof that P Q will be left as an exercise. To prove
that Q = P, we will prove the contrapositive.
Suppose lim f (x) # L. By the negation of the definition, there exists e>0
such that for each 8 > 0 there is a point x e I such that
0 < ]x – c| < 8 and f(x) – L| 2E.
In particular, for each positive integer n, there exists a point x, E I such that
0 < [xn - c| < 1/n and |f (x,) –L > E.
Now {Xn} is a sequence in I\ {c} that converges to c, but the sequence {f (xn)} does
not converge to L. This completes the proof of part (a). Part (b) of the theorem
follows from part (a); the details will be left as an exercise.
The argument found in the proof of part (a) of Theorem 3.2 is a common
one. The fact that something is true for each positive number is used to generate
a sequence. This is done by taking 1/n as the positive number for each positive
integer n and finding an appropriate x. Since this type of argument will be used
several times in this text, it is important to study it carefully.
Part (b) of Theorem 3.2 provides a condition for proving that a limit does not
exist that is often much easier to apply than the definition of a limit. As an example,
consider the function h defined by
if x is rational;
if x is irrational;
x,
h(x) =
%3D
-x,
d let c be any nonzero number. By previous results, there exists a sequence (x} of
Tational numbers such that r → c and x, c for all n, and there exists a sequence
PAT Or irrational numbers such that y, → c and y, #c for all n. Then](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde096970-4fbf-4bcb-afca-efa41531f77d%2Fecc18734-cfd0-412b-81d8-5d3f647552b8%2F1s4hkjh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:mathematically before reading the proof. (It might be helpful to read the discussion
that follows Theorem 2.4 or to look at Appendix A.)
THEOREM 3.2
THEIS a function that is defined on I except possibly at the point c.
Let I be an open interval that contains the point c and suppose
a) The function f has limit L at c if and only if for each sequence {Xn) in
I\ {c} that converges to c, the sequence (f (x,)} converges to L.
b) Suppose that there are sequences (x,} and (y,} in I\ {c} that converge
to c such that {ƒ(xn)} converges to L, and (f(yn)} converges to L2. If
Li + L2, then the functionf does not have a limit at c.
Proof Let P be the statement "the function f has limit L at c" and let Q be
the statement "for each sequence {xn} in I\ {c} that converges to c, the sequence
(f(xn)} converges to L". A proof that P Q will be left as an exercise. To prove
that Q = P, we will prove the contrapositive.
Suppose lim f (x) # L. By the negation of the definition, there exists e>0
such that for each 8 > 0 there is a point x e I such that
0 < ]x – c| < 8 and f(x) – L| 2E.
In particular, for each positive integer n, there exists a point x, E I such that
0 < [xn - c| < 1/n and |f (x,) –L > E.
Now {Xn} is a sequence in I\ {c} that converges to c, but the sequence {f (xn)} does
not converge to L. This completes the proof of part (a). Part (b) of the theorem
follows from part (a); the details will be left as an exercise.
The argument found in the proof of part (a) of Theorem 3.2 is a common
one. The fact that something is true for each positive number is used to generate
a sequence. This is done by taking 1/n as the positive number for each positive
integer n and finding an appropriate x. Since this type of argument will be used
several times in this text, it is important to study it carefully.
Part (b) of Theorem 3.2 provides a condition for proving that a limit does not
exist that is often much easier to apply than the definition of a limit. As an example,
consider the function h defined by
if x is rational;
if x is irrational;
x,
h(x) =
%3D
-x,
d let c be any nonzero number. By previous results, there exists a sequence (x} of
Tational numbers such that r → c and x, c for all n, and there exists a sequence
PAT Or irrational numbers such that y, → c and y, #c for all n. Then
Transcribed Image Text:112
II
2x +3
9. Use Theorem 3.2 to prove that lim
x2 x2 + 5x
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