The Limit Laws
In the following exercises (1-2), use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).
- Use constant multiple law and difference law: limx→0(4x2−2x+3)=4limx→0x2−2limx→0x+limx→03=3
- Use root law: limx→−2√x2−6x+3=√limx→−2(x2−6x+3)=√19
- 49
- 1
- −57
- limx→4x2−16x−4=16−164−4=00; then, limx→4x2−16x−4=limx→4(x+4)(x−4)x−4=8
- limx→63x−182x−12=18−1812−12=00; then, limx→63x−182x−12=limx→63(x−6)2(x−6)=32
- limt→9t−9√t−3=9−93−3=00; then, limt→9t−9√t−3=limt→9t−9√t−3√t+3√t+3=limt→9(√t+3)=6
- limθ→πsinθtanθ=sinπtanπ=00; then, limθ→πsinθtanθ=limθ→πsinθsinθcosθ=limθ→πcosθ=−1.
- limx→1/22x2+3x−22x−1=12+32−21−1=00; then, limx→1/22x2+3x−22x−1=limx→1/2(2x−1)(x+2)2x−1=52
- limx→62f(x)g(x)=2limx→6f(x)limx→6g(x)=72
- limx→6(f(x)+13g(x))=limx→6f(x)+13limx→6g(x)=7
- limx→6√g(x)−f(x)=√limx→6g(x)−limx→6f(x)=√5
- limx→6[(x+1)⋅f(x)]=(limx→6(x+1))(limx→6f(x))=28.
- 9
- 7
- 1
- 1
- limx→−3−(f(x)−3g(x))=limx→−3−f(x)−3limx→−3−g(x)=0+6=6
- limx→−52+g(x)f(x)=2+(limx→−5g(x))limx→−5f(x)=2+02=1
- limx→13√f(x)−g(x)=3√limx→1f(x)−limx→1g(x)=3√2+5=3√7
- limx→−9(x⋅f(x)+2g(x))=(limx→−9x)(limx→−9f(x))+2limx→−9(g(x))=(−9)(6)+2(4)=−46
Continuity
For the following exercises (1-4), determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.
- The function is defined for all x in the interval (0,∞).
- Removable discontinuity at x=0; infinite discontinuity at x=1
- f(x)=5ex−2 Infinite discontinuity at x=ln2
- H(x)=tan2x Infinite discontinuities at x=(2k+1)π4, for k=0,±1,±2,±3,⋯
- No. It is a removable discontinuity.
- Yes. It is continuous.
- Yes. It is continuous.
- k=−5
- k=−1
- k=163
It is not possible to redefine f(1) since the discontinuity is a jump discontinuity.
-
Answers may vary; see the following example:
-
Answers may vary; see the following example:
/li>
- False. It is continuous over (−∞,0)∪(0,∞).
- False. Consider f(x)={x if x≠04 if x=0
- False. Consider f(x)=cos(x) on [−π,2π].
- False. The IVT does not work in reverse! Consider (x−1)2 over the interval [−2,2].
- For all values of a,f(a) is defined, limθ→af(θ) exists, and limθ→af(θ)=f(a). Therefore, f(θ) is continuous everywhere.
- Nowhere
The Precise Definition of a Limit
In the following exercises (1-2), write the appropriate ε–δ definition for each of the given statements.
- For every ε>0, there exists a δ>0 so that if 0<|t−b|<δ, then |g(t)−M|<ε
- For every ε>0, there exists a δ>0 so that if 0<|x−a|<δ, then |ϕ(x)−A|<ε
- δ≤0.25
- δ≤2
- δ≤1
- δ<0.3900
- Let δ=ε. If 0<|x−3|<ε, then |x+3−6|=|x−3|<ε.
- Let δ=4√ε. If 0<|x|<4√ε, then |x4|=x4<ε.
- Let δ=ε2. If 5−ε2<x<5, then |√5−x|=√5−x<ε.
- Let δ=ε/5. If 1−ε/5<x<1, then |f(x)−3|=5x−5<ε.
- Let δ=√3N. If 0<|x+1|<√3N, then f(x)=3(x+1)2>N.
- f(x)−g(x)=f(x)+(−1)g(x)
- Answers may vary.