Introduction To Limits

By cbsemasterclasses


The concept/notation of limit is one of the most useful and fundamental ideas that distinguishes calculus from algebra and geometry. Even though the word limit is used in everyday language, the mathematical notion of a limit in calculus is completely different. To understand it clearly, 3 important questions need to be answered:

  • Limit of what? → Functions!
  • Why evaluate limit?
  • How to evaluate limit?

Why Do We Study Limits ?

In physical sciences and engineering, we sometimes encounter functions (representing physical quantities) that are undefined at x = a because they take an indeterminate form. However, their values as x → a are significant — e.g., instantaneous speed, acceleration, etc.

Hence, the need to compute:
limₓ→ₐ f(x)

Seven Indeterminate Forms- 0/0, ∞/∞, ∞ − ∞, 0 × ∞, 1∞, 00, ∞0

Five Fundamental Theorems (Limit Laws)

If: limₓ→c f(x) = L,     limₓ→c g(x) = M

Then:

1. Sum Rule: limₓ→c [f(x) + g(x)] = L + M

2. Difference Rule: limₓ→c [f(x) − g(x)] = L − M

3. Product Rule: limₓ→c [f(x) × g(x)] = L × M

4. Quotient Rule: limₓ→c f(x)/g(x) = L/M,   M ≠ 0

5. Constant Multiple Rule: limₓ→c K·f(x) = K·L

Strategies to Evaluate Limits

  • Factorisation
  • Rationalisation / Double rationalisation
  • Binomial theorem
  • Algebraic identities
  • “Law of Love” technique: We love ∞ in the denominator and 0 in the numerator

Examples

  1. limₓ→π/4 [(cot x − 1)/(2 cot x − 3)] = 3/4
  2. limₓ→0 [(5x − √(5x))/x] = √5
  3. limₓ→π/2 [tan(2x) + sin(3x) + sin(4x) + sin(6x)] = 1/12
  4. limₓ→9 [(4x − 2x² − 3)/(x − 3)] = 2/3
  5. limₓ→1 [(2x − 5)/(x² − 1)] = 1/2
  6. limₙ→∞ [(6n² + 2n − 1)/(3n³ + 2n² + 4n + 5)] = 1
  7. limₓ→∞ [(3x − 1)(5x + 1)/(2x − 40)] = 3/2
  8. limₓ→2 [(3x² − x − 6)/(x² − 4x + 4)] = 8
  • Step function: If f(x) = [x], then limₓ→5⁺ f(x) = 1, limₓ→5⁻ f(x) = 0
  • Modulus: limₓ→3 |x − 3| / (x² − 9) = D.N.E.

Trigonometric Limit Theorems

Theorem 1- If x is small and in radians:

1. limₓ→0 (sin x)/x = 1

2. limₓ→0 x/sin x = 1

3. limₓ→0 x·cosec x = 1

4. limₓ→0 (1 − cos x)/x = 0

Sandwich (Squeeze) Theorem

If h(x) ≤ f(x) ≤ g(x) and both limₓ→c h(x) = limₓ→c g(x) = L, then limₓ→c f(x) = L

Examples:

  • If √(25 − x²) < f(x) < √(5 − x²), then limₓ→0 f(x) = 5
  • 1 − x²/6 < 2(1 − cos x)/(x sin x) < 1 ⇒ limₓ→0 2(1 − cos x)/(x sin x) = 1
  • limₙ→∞ [n/(n + 1) + n/(n + 2) + … + n/(n + n)] = 1

Lecture 3: Exponential Functions, Theorem:

1. limx→0 (ax − 1)/x = ln a     (for a > 0)
2 .limx→0 (ex − 1)/x = 1

Examples on Exponential Functions:

1. limx→0 (etan x − 1)/x = 4

2. limx→∞ x(1 − e1/x) (Ans: 1)

3. limx→0 (ex tan x − 1)/x tan x = 1

4. limx→0 (ex cos x − 1)/x = 1

Example 4: limh→0 [(a + h)x − ax]/h = ax ln a

Example 5: limx→0 [(ex − cos x)/x] = −2

Example 6: limx→−∞ e1 / tan⁻¹(x − 1) = 1

Indeterminate Form 1∞

Theorem: limx→0 (1 + x)1/x = e
More generally: limn→∞ (1 + 1/n)n = e

Alternate Limit: limx→0 (1 + f(x))g(x) = elim f(x)·g(x)

Examples:- 1. limx→0 (1 + 3x)1/x = e³

2. limx→0 (1 + 5x)2/x = e10

3. limx→∞ (1 + K/x)xm = eKm

4. limx→e [(xn − 1)/(x − 1)] = n·en−1

5. limx→1 [(log₂ x)/(x − 1)] = 1/ln 2

6. limx→a [(ax − xa)/(x − a)] = ln a · aa

7. limx→a [(ax − xa)/(x − a)] = ln a · aa

Generalized Formula for 1∞: If: limx→a f(x) = 1 and limx→a φ(x) = ∞, then: limx→a [f(x)]φ(x) = elim (f(x) − 1)·φ(x)

Case: Finite Exponents If limx→a f(x) = A (A > 0), and limx→a φ(x) = B (finite), then: limx→a [f(x)]φ(x) = AB

Some More Examples: 1. limx→0 [(1 + x)1/x − e] = 0

2. limx→∞ [(1 + 1/x)x] = e

3. limx→0 [(ex − 1)/x] = 1

4. limx→0 [tan 2x / sin x] = 2

5. If f(x) = (sin x)(tan x) / (x²), evaluate limx→0 f(x) = 1

Limit Questions

1. limₓ→π (tan²x) / (cot(3x) + 2x) [Ans: e³]

2. limₓ→∞ (1 + 1/x)ˣ [Ans: e⁻¹]

3. limₓ→0 (1 + sinx + cos(ecx)) / tanx [Ans: 1]

4. limₓ→∞ (cosx / (sinx + 1/x)) [Ans: e]


5.(a) limₓ→0 (cos(ecx) / (x² + 9x + 5)) 

Options: (A) e⁻¹⁄⁶, (B) e⁻¹⁄⁵, (C) e⁻¹⁄²⁵, (D*) e⁻¹⁄³⁰

(b) limₓ→0 (cos(eˣ) × cot(2/x)) = ?

6.limₙ→∞ [cos(x/n)]ⁿ [Ans: e^(–x²/2)]

7. limₓ→0 tan⁴(x/π + 1) [Ans: e²]

8. limₓ→0 [cosᶰ(mx²)] m, n ∈ ℕ [Ans: e⁻²mn]

9. limₙ→∞ (a – 1 + b/n)ⁿ (a > 0, b > 0, n ∈ ℕ) [Ans: b¹⁄ᵃ]

10. limₓ→∞ (1 + 1/x + 1/(2x) + … + 1/(nx))ˣ [Ans: n!]

11. limₓ→0 [sin(ax² – π) / sec²(bx² – π)] = e⁻ᵃ²⁄ᵇ²

[Solution uses cosine expansion: limₓ→0 (cos(ax²–π) / cos(bx²–π)) = e⁻ᵃ²⁄ᵇ²]

12. limₓ→∞ [sin(n·sinx) / cos²x], with substitution x = π/t [Ans: –2π⁄4]

    Limits of Functions with Built-in Limits

    1. f(x) = [sin(x+1)] / [tan(x+π)] + 1/(n² + x²)

    limₓ→0⁺ f(x) = 0; limₓ→0⁻ f(x) = π ⇒ limit does not exist

    2. f(x) = [cos(x–1)] / [sin(x–1)] – 1/(n² + 1)

    limₓ→1 f(x) = –1

      Standard and Special Limits

      1.limₓ→0 (sinx)/x = 1

      2. limₓ→0 (cosx – 1)/x² = –½

      3. limₓ→0 (tanx)/x = 1

      4. limₓ→0 (ln(1 + x))/x = 1

      5. limₓ→0 (eˣ – 1)/x = 1

      6. limₓ→∞ (sinx)/x = 0

      Advanced Composite Limits

      • limₓ→0 (x – sinx)/x³ = 1⁄6
      • limₓ→0 (eˣ – 1 – x)/x² = ½
      • limₓ→0 (tanx – x)/x³ = 1⁄3

      Piecewise Function Behavior

      Let f(x) = { xsinx if x ≠ 0, 0 if x = 0 }
      Then: limₓ→0 f(x) = 0

      Other Important Limits

      • limₓ→∞ (cosx + x) / (sinx + x) = 1
      • limₓ→∞ ln(2x)/eˣ = 0
      • limₙ→∞ 2n / n! = 0
      • limₓ→0 (1 – cosx)/x² = ½
      • limₓ→0 log₂(1 + x)/x = log₂e

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