Prove that lim V = Va if a > 0. Hint: Use IVx - Val = Given e > 0, we must find 8---Select-- - such that IVx - Val -Select.-- whenever 0 < |x - al < ? . But IVx - Vāl = < ɛ (from the hint). Now if we can find a positive Vx + Va constant C such that Vx + Va ? C then Ix - a < I - al, and we can make X= al < ɛ by taking |x - al < Cɛ. We can find this number by restricting x to lie in some interval C C centered at a. If |x - al < 글이 then - Vx + va 2 V + Vā, and so C- Vta + Va is a suitable choice for the constant. So |x - al < + V This suggests that we let 8 - min + V. Thus, if 0 < |x - al < ? , then |x - al < | < + va ɛ, sO (Va/2 + Vā)e - E. (Va/2 + Va) IVx - val - Therefore, lim Vx = Va by the definition of a limit.
Prove that lim V = Va if a > 0. Hint: Use IVx - Val = Given e > 0, we must find 8---Select-- - such that IVx - Val -Select.-- whenever 0 < |x - al < ? . But IVx - Vāl = < ɛ (from the hint). Now if we can find a positive Vx + Va constant C such that Vx + Va ? C then Ix - a < I - al, and we can make X= al < ɛ by taking |x - al < Cɛ. We can find this number by restricting x to lie in some interval C C centered at a. If |x - al < 글이 then - Vx + va 2 V + Vā, and so C- Vta + Va is a suitable choice for the constant. So |x - al < + V This suggests that we let 8 - min + V. Thus, if 0 < |x - al < ? , then |x - al < | < + va ɛ, sO (Va/2 + Vā)e - E. (Va/2 + Va) IVx - val - Therefore, lim Vx = Va by the definition of a limit.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Prac. 13
Transcribed Image Text:|x - a||
Hint: Use IVx - val = Tx+ Va
Prove that lim Vx = Va if a > 0.
|x - a|
Given ɛ > 0, we must find 8 ---Select--
such that IVx - Val
Select--
whenever 0 < |x - al <? v
But Vx - Va| =
Vx + Va
< ɛ (from the hint). Now if we can find a positive
constant C such that Vx +
C then
|x – a|
<X = al, and we can make Xa ɛ by taking |x – al < Cɛ. We can find this number by restricting x to lie in some interval
centered at a. If |x – al < a, then
a <
Vx +
<X <
+
and so C =
Va + va is a suitable choice for the constant. So |x – al <
+ V.
Je. This suggests that we let 8 = mina,
Thus, if 0 < |x – a| < ? , then
+
|x - al <a
Vx + Va 2 Va + Va.
Also |x - a| <
+ va Je, so
|x - a|
Vx +
(Va/2 + Vā)ɛ
= E.
(Va/2 + Va)
IVx - Val =
Therefore, lim Vx = Va by the definition of a limit.
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