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Limits of Functions

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Limits of Functions

🎯 Learning Objectives

By the end of this topic, you will be able to:
  1. Define the limit of a function at a point
  2. Compute one-sided limits and determine when a two-sided limit exists
  3. Evaluate limits using algebraic manipulation, factoring, rationalization
  4. Apply standard limit results: sin⁡xx\frac{\sin x}{x}, ex−1x\frac{e^x-1}{x}, ln⁡(1+x)x\frac{\ln(1+x)}{x}
  5. Recognize and resolve indeterminate forms using L'Hôpital's rule

📋 Prerequisites


📖 Core Content

11.1 Intuition: Approaching a Point

What happens to f(x)=x2f(x) = x^2 as xx gets closer and closer to 33? The output approaches 99. We write lim⁡x→3x2=9\lim_{x \to 3} x^2 = 9. The limit describes the trend of f(x)f(x) near x=ax = a, regardless of whether f(a)f(a) is defined. This is crucial for understanding derivatives and continuity.
💡 Why this matters: Limits are the foundation of calculus. Derivatives = limits of difference quotients. Integrals = limits of Riemann sums. Everything in calculus rests on limits.

11.2 Definition

lim⁡x→af(x)=L\lim_{x \to a} f(x) = L
Means: as xx gets arbitrarily close to aa (from either side), f(x)f(x) gets arbitrarily close to LL.

11.2.1 One-Sided Limits

  • Left-hand limit: lim⁡x→a−f(x)\lim_{x \to a^-} f(x) — approach aa from below (smaller values)
  • Right-hand limit: lim⁡x→a+f(x)\lim_{x \to a^+} f(x) — approach aa from above (larger values) Two-sided limit exists iff both one-sided limits exist and are equal:
lim⁡x→af(x)=L  ⟺  lim⁡x→a−f(x)=lim⁡x→a+f(x)=L\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L
(Diagram)

11.3 Properties of Limits

If lim⁡x→af(x)=L\lim_{x \to a} f(x) = L and lim⁡x→ag(x)=M\lim_{x \to a} g(x) = M:
PropertyFormula
Sumlim⁡[f(x)+g(x)]=L+M\lim[f(x) + g(x)] = L + M
Differencelim⁡[f(x)−g(x)]=L−M\lim[f(x) - g(x)] = L - M
Productlim⁡[f(x)⋅g(x)]=L⋅M\lim[f(x) \cdot g(x)] = L \cdot M
Quotientlim⁡f(x)/g(x)=L/M\lim f(x)/g(x) = L/M (if M≠0M \neq 0 )
Constant multiplelim⁡c⋅f(x)=cL\lim c \cdot f(x) = cL
Powerlim⁡[f(x)]n=Ln\lim [f(x)]^n = L^n
Rootlim⁡f(x)n=Ln\lim \sqrt[n]{f(x)} = \sqrt[n]{L} (if L≥0L \geq 0 for even nn )

11.4 Standard Limits (Very Important!)

FormulaNotes
lim⁡x→0sin⁡xx=1\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1xx in radians
lim⁡x→01−cos⁡xx=0\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = 0
lim⁡x→0ex−1x=1\displaystyle \lim_{x \to 0} \frac{e^x - 1}{x} = 1
lim⁡x→0ln⁡(1+x)x=1\displaystyle \lim_{x \to 0} \frac{\ln(1+x)}{x} = 1
lim⁡x→∞(1+1x)x=e\displaystyle \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e
lim⁡x→0(1+x)1/x=e\displaystyle \lim_{x \to 0} (1 + x)^{1/x} = eSame as above

11.5 Evaluation Techniques

11.5.1 Direct Substitution

If ff is continuous at aa, lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a). Works for polynomials, rational functions (where denominator ≠ 0), trig functions, exponentials.

11.5.2 Factoring

When substitution gives 0/00/0, factor and cancel. Example: lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}
lim⁡x→2(x−2)(x+2)x−2=lim⁡x→2(x+2)=4\lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4

11.5.3 Rationalization

When radicals cause 0/00/0, multiply by conjugate. Example: lim⁡x→0x+1−1x\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}
lim⁡x→0(x+1−1)(x+1+1)x(x+1+1)=lim⁡x→0xx(x+1+1)=lim⁡x→01x+1+1=12\lim_{x \to 0} \frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{x(\sqrt{x+1}+1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+1}+1)} = \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{2}

11.5.4 Limits at Infinity

Divide numerator and denominator by highest power of xx. Example: lim⁡x→∞3x2+2x4x2−1=34\lim_{x \to \infty} \frac{3x^2 + 2x}{4x^2 - 1} = \frac{3}{4}

11.6 Indeterminate Forms

FormExample
0/00/0x2−4x−2\frac{x^2-4}{x-2} at x=2x=2
∞/∞\infty/\inftyx2+1x−1\frac{x^2+1}{x-1} as x→∞x\to\infty
0⋅∞0 \cdot \inftyx⋅ln⁡xx \cdot \ln x as x→0+x\to 0^+
∞−∞\infty - \inftyx+1−x\sqrt{x+1} - \sqrt{x} as x→∞x\to\infty
000^0xxx^x as x→0+x\to 0^+
1∞1^\infty(1+1/x)x(1+1/x)^x as x→∞x\to\infty
∞0\infty^0x1/xx^{1/x} as x→∞x\to\infty

11.7 L'Hôpital's Rule

If lim⁡x→af(x)/g(x)\lim_{x \to a} f(x)/g(x) is 0/00/0 or ∞/∞\infty/\infty, and lim⁡f′(x)/g′(x)\lim f'(x)/g'(x) exists:
lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}
Important: Only applies to 0/00/0 or ∞/∞\infty/\infty forms. Must differentiate numerator and denominator separately (not quotient rule!).

Worked Examples

Example 4.1: lim⁡x→0sin⁡xx\lim_{x \to 0} \frac{\sin x}{x} (using L'Hôpital)
lim⁡x→0sin⁡xx=lim⁡x→0cos⁡x1=1\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1
Example 4.2: lim⁡x→∞x2ex\lim_{x \to \infty} \frac{x^2}{e^x}
lim⁡x→∞x2ex=lim⁡x→∞2xex=lim⁡x→∞2ex=0\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} = 0
Example 4.3: lim⁡x→0ex−1−xx2\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}
lim⁡x→0ex−1−xx2=lim⁡x→0ex−12x=lim⁡x→0ex2=12\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}

📐 Key Formulas — Summary Table

FormTechnique
0/00/0 (polynomial)Factor and cancel
0/00/0 (radical)Rationalize
∞/∞\infty/\inftyDivide by highest power
0/00/0 or ∞/∞\infty/\inftyL'Hôpital's rule
0⋅∞0 \cdot \inftyRewrite as 0/00/0 or ∞/∞\infty/\infty
∞−∞\infty - \inftyCommon denominator or conjugate
1∞1^\inftyUse lim⁡=elim⁡(f(x)−1)g(x)\lim = e^{\lim (f(x)-1)g(x)}
Standard limit: lim⁡x→0sin⁡xx=1, lim⁡x→0ex−1x=1, lim⁡x→0ln⁡(1+x)x=1\boxed{\text{Standard limit: } \lim_{x\to 0} \frac{\sin x}{x} = 1,\ \lim_{x\to 0} \frac{e^x-1}{x} = 1,\ \lim_{x\to 0} \frac{\ln(1+x)}{x} = 1}

⚠️ Common Pitfalls

Pitfall 1: Applying L'Hôpital to Non-Indeterminate Forms

lim⁡x→1x2x=1\lim_{x \to 1} \frac{x^2}{x} = 1 directly. Using L'Hôpital: 2x1=2\frac{2x}{1} = 2 — WRONG! The form was 1/11/1, not 0/00/0 or ∞/∞\infty/\infty.

Pitfall 2: Confusing One-Sided Limits with Two-Sided

If lim⁡x→a−f(x)≠lim⁡x→a+f(x)\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x), then lim⁡x→af(x)\lim_{x \to a} f(x) does NOT exist.

Pitfall 3: Forgetting That sin⁡x/x\sin x/x Requires Radians

The limit lim⁡x→0sin⁡xx=1\lim_{x\to 0} \frac{\sin x}{x} = 1 only holds when xx is in radians.

📝 Practice Questions

Q1: Find lim⁡x→3(2x+1)\lim_{x \to 3} (2x + 1).
Direct substitution: 2(3)+1=72(3) + 1 = 7
7\boxed{7} Q2: Find lim⁡x→2x2−3x+2x−2\lim_{x \to 2} \frac{x^2 - 3x + 2}{x-2}.
Factor: (x−1)(x−2)x−2=x−1→1\frac{(x-1)(x-2)}{x-2} = x-1 \to 1
1\boxed{1} Q3: Find lim⁡x→04+x−2x\lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x}.
Rationalize: (4+x−2)(4+x+2)x(4+x+2)=xx(4+x+2)=14+x+2→14\frac{(\sqrt{4+x} - 2)(\sqrt{4+x}+2)}{x(\sqrt{4+x}+2)} = \frac{x}{x(\sqrt{4+x}+2)} = \frac{1}{\sqrt{4+x}+2} \to \frac{1}{4}
14\boxed{\frac{1}{4}} Q4: Find lim⁡x→∞5x2−3x+12x2+7\lim_{x \to \infty} \frac{5x^2 - 3x + 1}{2x^2 + 7}.
Divide by x2x^2: 5−3/x+1/x22+7/x2→52\frac{5 - 3/x + 1/x^2}{2 + 7/x^2} \to \frac{5}{2}
52\boxed{\frac{5}{2}} Q5: Find lim⁡x→0sin⁡3xx\lim_{x \to 0} \frac{\sin 3x}{x}.
sin⁡3xx=3⋅sin⁡3x3x→3⋅1=3\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x} \to 3 \cdot 1 = 3
3\boxed{3} Q6: Find lim⁡x→01−cos⁡xx2\lim_{x \to 0} \frac{1 - \cos x}{x^2}.
=lim⁡x→0sin⁡x2x=12⋅1=12= \lim_{x\to 0} \frac{\sin x}{2x} = \frac{1}{2} \cdot 1 = \frac{1}{2}
12\boxed{\frac{1}{2}} Q7: Find lim⁡x→0e2x−1x\lim_{x \to 0} \frac{e^{2x} - 1}{x}.
=2⋅lim⁡x→0e2x−12x=2⋅1=2= 2 \cdot \lim_{x\to 0} \frac{e^{2x} - 1}{2x} = 2 \cdot 1 = 2
2\boxed{2} Q8: Evaluate lim⁡x→0x2sin⁡x\lim_{x \to 0} \frac{x^2}{\sin x}.
=lim⁡x→0xsin⁡x⋅x=1⋅0=0= \lim_{x\to 0} \frac{x}{\sin x} \cdot x = 1 \cdot 0 = 0
0\boxed{0} Q9: Find lim⁡x→1x3−1x−1\lim_{x \to 1} \frac{x^3 - 1}{x - 1}.
Factor: (x−1)(x2+x+1)x−1=x2+x+1→3\frac{(x-1)(x^2+x+1)}{x-1} = x^2 + x + 1 \to 3
3\boxed{3} Q10: Find lim⁡x→0+xln⁡x\lim_{x \to 0^+} x \ln x.
Rewrite as ln⁡x1/x\frac{\ln x}{1/x}: −∞/∞-\infty/\infty form. L'Hôpital: 1/x−1/x2=−x→0\frac{1/x}{-1/x^2} = -x \to 0
0\boxed{0} Q11: Find lim⁡x→0ex−1sin⁡x\lim_{x \to 0} \frac{e^x - 1}{\sin x}.
L'Hôpital: excos⁡x→11=1\frac{e^x}{\cos x} \to \frac{1}{1} = 1
1\boxed{1} Q12: Find lim⁡x→0(1+2x)1/x\lim_{x \to 0} (1 + 2x)^{1/x}.
This is 1∞1^\infty form. lim⁡x→0(1+2x)1/x=elim⁡x→0ln⁡(1+2x)x=elim⁡x→021+2x=e2\lim_{x\to 0} (1+2x)^{1/x} = e^{\lim_{x\to 0} \frac{\ln(1+2x)}{x}} = e^{\lim_{x\to 0} \frac{2}{1+2x}} = e^{2}
e2\boxed{e^2}

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