Chapter 1: Limits
Section 1.2: Precise Definition of a Limit
Example 1.2.5
Use Definition 1.2.1 to verify limx→3(3⁢x−4)=5, the limit explored in Example 1.2.1.
Solution
Type the equation fx=…
Context Panel: Assign Function
fx=3 x−4→assign as functionf
Figure 1.2.5(a) is an animation in which y=5 is graphed in blue, and fx=3 x−4, 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=5 , and red and green vertical lines delineate a corresponding δ-band around x=3.
The red and green horizontal lines are drawn at y=5 ±ϵ, respectively, and the red and green vertical lines are drawn at the corresponding x-coordinates x=3 ±δ=f−15 ±ϵ. (Because of the linearity of f, δL=δR=δϵ).
Figure 1.2.5(a) Animation illustrating Definition 1.2.1
Write the equation fa−δL=L−ϵ Press the Enter key.
Context Panel: Solve≻Isolate Expression for≻δL
f3− δL=5− ϵ
5−3⁢δL=5−ϵ
→isolate for delta[L]
δL=13⁢ϵ
Write the equation fa+δR=L−ϵ Press the Enter key.
Context Panel: Solve≻Isolate Expression for≻δR
f3+δR=5+ϵ
5+3⁢δR=5+ϵ
→isolate for delta[R]
δR=13⁢ϵ
Consequently, δL=δR=δϵ=ϵ/3. To complete the proof, show that x−3<ϵ/3 ⇒ 3 x−4−5<ϵ. This is done by showing that f3+t ϵ/3−5<ϵ, where t<1.
|f3+t ϵ/3−5| = t ϵ = t ϵ = t ϵ<ϵ
The first equality follows from straight-forward algebra, and the second 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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