Chapter 3 – Limit Laws and Algebraic Techniques

3.1 Learning Objectives

By the end of this chapter, you should be able to:

• Apply limit laws to build complex limits from simpler ones.
• Evaluate limits of polynomials and rational functions using direct substitution when possible.
• Recognize indeterminate forms like 0/0 and know you must transform the expression.
• Use factoring and cancellation to resolve 0/0 forms.
• Use conjugates to handle limits with square roots.
• Evaluate basic trigonometric limits such as limx→0​xsinx​ and limx→0​x21−cosx​.
• Apply the Squeeze Theorem in simple situations.
• Distinguish between limits that exist and are finite, infinite limits, and limits that do not exist.

3.2 Big Picture: From Intuition to Computation

In Chapter 2 you learned to see limits: by tables, graphs, and intuition (“what value does the function approach?”). That gave you a conceptual sense of what a limit is.

Now we focus on how to actually compute limits by hand. The goals:

• Turn a messy expression into something you can evaluate safely by substitution.
• Have a decision process: what should you try first, second, third?
• Learn a handful of algebraic “moves” that resolve the most common problem types.

From here on, limits are not just ideas; they become tools you can use to define derivatives and integrals.

3.3 Limit Laws (How and When to Use Them)

3.3.1 The idea

Limit laws tell you that limits respect the usual arithmetic operations, as long as the individual limits exist and you don’t divide by 0.

If you know:

x→alim​f(x)= L,x→alim​g(x)= M,

then you automatically know:

• Sum/difference: x→alim​[f(x)±g(x)]= L±M.
• Constant multiple: x→alim​[cf(x)]= cL.
• Product: x→alim​[f(x)g(x)]= LM.
• Quotient (if M=0): x→alim​g(x)f(x)​= ML​.

Interpretation:

You can safely “pull apart” complicated expressions into pieces, evaluate each piece’s limit, then rebuild.

Tutor process cue:

1. Break the expression into simpler pieces (sums, products, etc.).
2. Take limits of each piece (often just by plugging in).
3. Reassemble using the laws.

3.3.2 Polynomials and rational functions

For a polynomial p(x):

x→alim​p(x)= p(a).

So “the limit is what you get by plugging a into the polynomial.” No tricks needed.

For a rational function q(x)p(x)​:

• If q(a)=0, then

x→alim​q(x)p(x)​= q(a)p(a)​.

You can just substitute directly.

• If q(a)=0, then you have a potential problem (0 in the denominator).

You must:

• either simplify first (often leading to a removable 0/0), or
• recognize that the limit might not be finite (vertical asymptote).

This gives a first step in your decision process:

Step 1: Try direct substitution (especially for polynomials & rational functions).

- If you get a nice finite number → done.

- If you get 0/0 or something like 5/0 → switch to algebraic techniques.

---

3.4 Indeterminate Forms and Algebraic Techniques

3.4.1 Recognizing 0/0 (indeterminate form)

If plugging in x = a gives

• numerator → 0
• denominator → 0

you cannot conclude the limit is 0/0 (which is “nothing”). 0/0 is a warning label: the limit might be:

• a finite number,
• infinity,
• or not exist at all.

The only way to know is to simplify the expression first.

Common pattern: a factor of (x − a) appears in both numerator and denominator.

3.4.2 Process: factoring and canceling

Typical form:

x→alim​polynomialpolynomial​.

Process:

1. Plug in x = a.

• If not 0/0, you’re done.

1. If 0/0 occurs, factor numerator and denominator.
2. Look for a common factor (often (x − a) or (x + something)).
3. Cancel the common factor (this removes the “hole”).
4. Now try substitution again.

Key idea:

• When we cancel (x − a), we’re not changing the behavior near a, only at the single point x = a. That’s fine, because the limit cares only about behavior near a, not at a.

Tutor cue:

• Ask: “What common factor causes the 0? Can we factor it out and cancel?”

3.4.3 Process: rationalizing with conjugates

When you see square roots in the numerator or denominator and substitution gives 0/0, try rationalizing.

Typical forms:

• small denominator⋅
• ​−number​
• small denominatornumber−⋅
• ​​

Process:

1. Identify the “square root part” in the numerator (e.g. x+1

• ​−1).
• Multiply numerator and denominator by its conjugate:
• Conjugate of A

​−B​ is A​+B

1. - ​.
2. The numerator becomes a difference of squares: A−B, which often cancels the x in the denominator.
3. Simplify and then substitute.

Reason:

• The conjugate trick removes the radical from the numerator, exposing a factor that cancels the troublesome x or (x − a).

3.4.4 Trigonometric simplifications (using known limits)

Some limits are hard purely algebraically but easy with known standard limits:

• x→0lim​xsinx​=1.
• x→0lim​x21−cosx​=21​.

Near 0, you can think:

• sinx≈x,
• tanx≈x.

So:

• x→0lim​kxsin(kx)​=1,
• x→0lim​kxtan(kx)​=1,

and then scale appropriately.

Process pattern:

• Rewrite expressions like xsin(3x)​ into a form involving sin(3x)/(3x) times a constant.
• Rewrite expressions like x21−cosx​ into a known standard limit.

3.5 The Squeeze Theorem (When to Use It)

The Squeeze Theorem is useful when:

• You cannot simplify f(x) algebraically easily, but
• You can find two simple functions g(x) and h(x) such that g(x)≤f(x)≤h(x) near x = a, and both g and h have the same limit at a.

Then:

x→alim​g(x)= x→alim​h(x)= L⇒x→alim​f(x)= L.

Process:

1. Find inequalities bounding the tricky function (often using |sin|≤1, |cos|≤1).
2. Multiply or modify those inequalities to include x or x² factors.
3. Take the limits of the outer functions.
4. Conclude the middle function’s limit.

Classic moves:

• Use ∣sint∣≤1 to bound expressions like x·sin(1/x) or x²·cos(1/x²).
• Use −1≤cost≤1 in similar ways.

Tutor cue:

• Ask: “Can I trap this wild function between two simpler functions whose limits I know?”

3.6 Worked Mini-Examples (Step-by-step)

Example 1 – Polynomial (direct substitution)

Compute:

x→2lim​(3x2−5x+1).

Process:

1. Recognize a polynomial → limits by substitution.
2. Plug in x = 2:

3(2)2−5(2)+1=12−10+1=3.

Example 2 – Rational, non-0 denominator

Compute:

x→1lim​x2+12x2−x+3​.

1. Check denominator at x = 1: 12+1=2=0 → substitution okay.
2. Plug in:

12+12(1)2−1+3​=22−1+3​=24​=2.

Example 3 – Rational, 0/0 → factor & cancel

Compute:

x→3lim​x−3x2−9​.

1. Substitute: numerator: 9 − 9 = 0; denominator: 3 − 3 = 0 → 0/0 indeterminate.
2. Factor numerator: x2−9=(x−3)(x+3).
3. Cancel:

x−3(x−3)(x+3)​= x+3(x=3).

1. Now limit of x+3 as x→3 is 6.

Example 4 – Radical → conjugate

Compute:

x→0lim​xx+1​−1​.

1. Substitute: numerator: √1 − 1 = 0; denominator: 0 → 0/0.
2. Multiply by conjugate x+1

​+1:

xx+1​−1​⋅x+1​+1x+1​+1​= x(x+1​+1)(x+1)−1​= x(x+1​+1)x​= x+1

1. ​+11​.
2. Substitute x = 0:

1+11​=21​.

3.7 Practice Problems (30 Problems)

(Same problems as before; only the explanations above are richer.)

A. Direct Substitution and Basic Limit Laws (1–8)

1. Evaluate:

x→2lim​(5x2−3x+4).

1. Evaluate:

x→−1lim​(x3+4x2−x+7).

1. Evaluate:

x→0lim​(7−2x+x2).

1. Evaluate:

x→3lim​x2+2x−34x2−x+1​.

1. Evaluate:

x→1lim​x2−1x2+3x+2​.

1. Evaluate:

x→−2lim​x2+42x2+x−3​.

1. Evaluate:

x→4lim​(3−x2​).

1. Suppose limx→2​f(x)=5 and limx→2​g(x)=−3. Find:

a) limx→2​[2f(x)−3g(x)]

b) limx→2​[f(x)g(x)]

c) limx→2​g(x)f(x)​

B. 0/0 Forms: Factor and Cancel (9–16)

1. x→2lim​x−2x2−4​
2. x→−1lim​x+1x2−1​
3. x→3lim​x−3x2−5x+6​
4. x→1lim​x−1x2−3x+2​
5. x→0lim​x2x3−2x2​
6. x→4lim​x−4x2−16​
7. x→5lim​x−5x2−25​
8. x→2lim​x−2x3−8​

C. Conjugates and Square Roots (17–22)

1. x→0lim​xx+4

• ​−2​
• x→5lim​x−5x

​−5- ​​

• x→0lim​x1−1−x
• ​​
• x→0lim​x9+x
• ​−3​
• x→0lim​1+x
• ​−1x​
• x→0lim​x1+2x

​−1−2x

1. ​​

D. Trigonometric Limits (23–26)

1. x→0lim​xsinx​
2. x→0lim​xsin(3x)​
3. x→0lim​x21−cosx​
4. x→0lim​xtan(2x)​

E. Squeeze Theorem and Nonexistent Limits (27–30)

1. Show, using the Squeeze Theorem, that:

x→0lim​x2cos(x1​)=0.

1. Explain why:

x→0lim​sin(x1​)

does not exist.

1. Evaluate:

x→0lim​xsin(x1​).

1. Decide whether this limit exists, and if so, find it:

x→0lim​x∣x∣​.

3.8 Solutions

(Same as the previous Chapter 3; I’ll keep them verbatim here.)

Section A – Direct Substitution and Limit Laws

x→2lim​(5x2−3x+4)=5(2)2−3(2)+4=20−6+4=18.

x→−1lim​(x3+4x2−x+7)=(−1)3+4(−1)2−(−1)+7=−1+4+1+7=11.

x→0lim​(7−2x+x2)=7−2(0)+02=7.

x→3lim​x2+2x−34x2−x+1​=32+2(3)−34(3)2−3+1​=9+6−336−3+1​=1234​=617​.

1. Denominator → 0, simplifies to (x+2)/(x−1). As x→1, denominator 0 ⇒

limit does not exist as a finite number (vertical asymptote).

x→−2lim​x2+42x2+x−3​=(−2)2+42(−2)2+(−2)−3​=88−5​=83​.

x→4lim​(3−x2​)=3−42​=25​.

8a) 19

8b) −15

8c) −5/3

Section B – 0/0 Forms

1. 4
2. −2
3. 1
4. −1
5. −2
6. 8
7. 10
8. 12

Section C – Conjugates

1. 1/4
2. 1/(2√5)
3. 1/2
4. 1/6
5. 2
6. 2

Section D – Trig Limits

1. 1
2. 3
3. 1/2
4. 2

Section E – Squeeze and Nonexistence

1. Squeezed between −x² and x² → limit 0.
2. Infinite oscillation, does not settle to one value → limit does not exist.
3. |x sin(1/x)| ≤ |x| → squeezed to 0 → limit 0.
4. Left limit −1, right limit 1 → no two-sided limit.