Chapter 1: Limits
Section 1.2: Precise Definition of a Limit
Example 1.2.6
Use Definition 1 to verify limx→3 x2−9x−3=6.
Solution
Type the equation fx=…
Context Panel: Assign Function
fx=x2−9x−3→assign as functionf
Figure 1.2.6(a) is an animation in which y=6 is graphed in blue, and fx=x2−9x−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=6 and red and green vertical lines delineate a corresponding δ-band around x=3.
Figure 1.2.6(a) Animation illustrating Definition 1.2.1
The red and green horizontal lines are drawn at y=6 ±ϵ, respectively, and the red and green vertical lines are drawn at the corresponding x-coordinates x=3 ±δ=f−16 ±ϵ. (Because f reduces to the linear x+3 for x≠3, once again δL=δR=δϵ).
Write the equation fa−δL=L− ϵ Press the Enter key.
Context Panel: Simplify≻Simplify
Context Panel: Solve≻Isolate Expression for≻δL
f3−δL=6− ϵ
−3−δL2−9δL=6−ϵ
= simplify
6−δL=6−ϵ
→isolate for delta[L]
δL=ε
Write the equation fa+δR=L+ϵ Press the Enter key.
Context Panel: Solve≻Isolate Expression for≻δR
f3+δR=6+ϵ
3+δR2−9δR=6+ϵ
6+δR=6+ϵ
→isolate for delta[R]
δR=ε
Consequently, δL=δR=δϵ=ϵ. To complete the proof, show that x−3<ϵ ⇒ fx−6<ϵ. This is done by showing that f3+t ϵ−6<ϵ, where t<1.
f3+t ϵ−6 = 3+t⁢ϵ2−9t⁢ϵ−6= simplify t⁢ϵ = t ϵ = t ϵ<ϵ
The first and second equalities follow from straight-forward algebra; the next, from the recognition that the absolute value of a product is the product of the absolute values. The next equality follows because ϵ is positive, and the final inequality follows because t<1.
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