Chapter 1: Limits
Section 1.2: Precise Definition of a Limit
Example 1.2.8
Use Definition 1.2.1 to verify limx→3x2−3x+3=3, the limit explored in Example 1.2.3.
Solution
Type the equation fx=…
Context Panel: Assign Function
fx=x2−3 x+3→assign as functionf
Figure 1.2.8(a) is an animation in which y=3 is graphed in blue, and fx=x2−3 x+3, in black.
The slider in the animation toolbar controls the value of ϵ. As the slider is moved past the first frame, red and green horizontal lines delineate an ϵ-band around y=3 and red and green vertical lines delineate the band 3−δL<x<3+δR.
The red and green horizontal lines are drawn at y=3 ±ϵ, respectively, and the red and green vertical lines are drawn at the corresponding x-coordinates 3−δL= f−13−ϵ and 3+δR= f−13+ϵ.
f:=x->x^2-3*x+3: g:=x->(3+sqrt(4*x-3))/2: p1:=plot(f,2..4,color=black): p2:=plots:-animate(plot,[[3+epsilon,3-epsilon,3],x=2..4,color=[green,red,blue]],epsilon=0..1,frames=26,background=p1): p3:=plots:-animate(plot,[[[g(3+epsilon),t*f(g(3+epsilon)),t=0..1],[g(3-epsilon),t*f(g(3-epsilon)),t=0..1]],color=[green,red],titlefont=[Times,14]],epsilon=0..1,frames=26): plots:-display(p||(1..3));
Figure 1.2.8(a) Animation illustrating Definition 1.2.1
Write the equation fa−δL=3−ϵ Press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻δL
f3−δL=3−ϵ
3−δL2−6+3δL=3−ϵ
→solutions for delta[L]
32+129−4ϵ,32−129−4ϵ
Select the solution that tends to 0 as ϵ→0. (The other solution tends to 3.) Hence, δ__L≔32−129−4 ϵ:
Write the equation fa+δR=3+ϵ Press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻δR
f3+δR=3+ϵ
3+δR2−6−3δR=3+ϵ
→solutions for delta[R]
−32+129+4ϵ,−32−129+4ϵ
Select the solution that tends to 0 as ϵ→0. (The other solution tends to −3.) Hence, δ__R≔−32+129+4ϵ:
Figure 1.2.8(b) provides graphical evidence that, for small values of ε, δR≤δL.
For an analytic determination that minδL,δR=δR, show δL/δR>1.
Begin by simplifying the ratio δL/δR , look at its graph, and finish via the calculations in Table 1.2.8(a).
use plots, plottools in module() local dL,dR,p1,p2,p3,ve; ve:=varepsilon; dL:=(3-sqrt(9-4*ve))/2; dR:=(sqrt(9+4*ve)-3)/2; p1:=plot([dR,dL],ve=0..1/2,color=[black,red],tickmarks=[5,3]); p2:=textplot({[.4,.18,typeset(delta[L])],[.4,.085,typeset(delta[R])]}); p3:=display(p1,p2,scaling=constrained); print(p3); end module: end use:
Figure 1.2.8(b)
δ__Lδ__R = 32−129−4ϵ−32+129+4ϵ= simplify −−3+9−4ϵ−3+9+4ϵ
The first step in Table 1.2.8(a) is the rationalization of the denominator; the second is making the resulting numerator smaller by changing 9+4 ϵ to 9−4 ϵ in the second factor.
δ__Lδ__R = 3−9−4 ϵ9+4 ϵ−3⋅9+4 ϵ+39+4 ϵ+3
= 3−9−4 ϵ3+9+4 ϵ4 ϵ
>3−9−4 ϵ3+9−4 ϵ4 ϵ
=4 ϵ4 ϵ=1
Table 1.2.8(a) Show δL/δR>1
Consequently, δϵ=δR=9+4 ϵ−3/2 . To complete the proof, show that x−3<δ=δR ⇒ fx−3<ϵ. This is done in Table 1.2.8(b) by showing that f3+t δR−3<ϵ, where t<1.
1
f3+t δ__R−3
=−92t+32t9+4ϵ+92t2−32t29+4ϵ+t2ϵ
2
=92−329+4ϵ+ϵt2+−92+329+4ϵt
3
≤|92−329+4ϵ+ϵ|t2+−92+329+4ϵ t
4
≤92−329+4ϵ+ϵ+−92+329+4ϵt
5
=ϵ t
6
<ϵ
Table 1.2.8(b) Verification that x=3+t δR ⇒ f3+t δR−3<ϵ
Lines (1) and (2) in Table 1.2.8(b) are obtained with basic algebra. The result in (3) is obtained by applying the triangle inequality: |a+b|≤a+b. The coefficients of t and t2 in (3) are positive (graph them!), so their absolute values are dropped in (4). Also in (4), replacing t2 with t makes the expression in (3) larger because for t≤1, t2≤t. Simple arithmetic applied to (4) gives (5), and hence (6).
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