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Last updated March 22, 2026

Basic Limit by Direct Substitution

The simplest limits are evaluated by substituting the value directly:

Problem: lim(x→3) of (x² + 2x - 1)

Step 1: Substitute x = 3
= (3)² + 2(3) - 1
= 9 + 6 - 1
= 14

Result: lim(x→3) (x² + 2x - 1) = 14

Direct substitution works when the function is continuous at the limit point and the denominator is not zero.

Limit Involving 0/0 Indeterminate Form — Factoring

When direct substitution gives 0/0, factor and simplify:

Problem: lim(x→2) of (x² - 4) / (x - 2)

Step 1: Direct substitution gives 0/0 (indeterminate)
  (4 - 4) / (2 - 2) = 0/0

Step 2: Factor the numerator
  (x² - 4) = (x + 2)(x - 2)

Step 3: Cancel the common factor (x - 2)
  (x + 2)(x - 2) / (x - 2) = x + 2

Step 4: Evaluate the simplified expression at x = 2
  x + 2 = 2 + 2 = 4

Result: lim(x→2) (x² - 4)/(x - 2) = 4

Applying L'Hôpital's Rule

When factoring doesn't work, L'Hôpital's rule differentiates numerator and denominator:

Problem: lim(x→0) of sin(x) / x

Step 1: Direct substitution gives 0/0 (indeterminate)

Step 2: Apply L'Hôpital's Rule
  Differentiate numerator: d/dx[sin(x)] = cos(x)
  Differentiate denominator: d/dx[x] = 1

Step 3: Evaluate the new limit
  lim(x→0) cos(x) / 1 = cos(0) / 1 = 1

Result: lim(x→0) sin(x)/x = 1

This is one of the most important limits in calculus, used in the derivation of the derivative of sin(x).

Limit at Infinity — Rational Function

For rational functions, the limit at infinity depends on the degrees of numerator and denominator:

Problem: lim(x→∞) of (3x² + 2x) / (5x² - 1)

Step 1: Divide numerator and denominator by x² (highest power)
  = (3 + 2/x) / (5 - 1/x²)

Step 2: As x→∞, terms with x in denominator approach 0
  2/x → 0
  1/x² → 0

Step 3: Evaluate
  = (3 + 0) / (5 - 0) = 3/5

Result: lim(x→∞) (3x² + 2x)/(5x² - 1) = 3/5

Horizontal asymptote: y = 3/5

Limit at Infinity — Exponential

Problem: lim(x→∞) of e^(-x)

As x increases, -x decreases without bound.
e^(-x) = 1/e^x → 0 as e^x → ∞

Result: lim(x→∞) e^(-x) = 0

Problem: lim(x→-∞) of e^(-x)

As x→-∞, -x→+∞
e^(-x) → ∞

Result: lim(x→-∞) e^(-x) = ∞

One-Sided Limits

Some functions have different left and right limits. The overall limit exists only if both sides agree:

Problem: lim(x→0) of |x| / x

Left-hand limit (x→0⁻, x is negative):
  |x| = -x when x < 0
  |x|/x = -x/x = -1
  lim(x→0⁻) = -1

Right-hand limit (x→0⁺, x is positive):
  |x| = x when x > 0
  |x|/x = x/x = 1
  lim(x→0⁺) = 1

Since left limit (-1) ≠ right limit (1):
Result: lim(x→0) |x|/x does NOT EXIST

Trigonometric Limit

Problem: lim(x→0) of (1 - cos(x)) / x²

Step 1: Direct substitution gives 0/0

Step 2: Apply L'Hôpital's Rule
  Numerator derivative: sin(x)
  Denominator derivative: 2x
  New limit: lim(x→0) sin(x) / (2x)

Step 3: Apply L'Hôpital's Rule again (still 0/0)
  Numerator derivative: cos(x)
  Denominator derivative: 2
  New limit: lim(x→0) cos(x) / 2

Step 4: Substitute x = 0
  cos(0) / 2 = 1/2

Result: lim(x→0) (1 - cos(x))/x² = 1/2

Limit Involving Infinity/Infinity — L'Hôpital's Rule

Problem: lim(x→∞) of ln(x) / x

Step 1: Direct substitution gives ∞/∞ (indeterminate)

Step 2: Apply L'Hôpital's Rule
  Numerator derivative: 1/x
  Denominator derivative: 1
  New limit: lim(x→∞) (1/x) / 1 = lim(x→∞) 1/x

Step 3: As x→∞, 1/x → 0

Result: lim(x→∞) ln(x)/x = 0

Interpretation: x grows faster than ln(x).
Even though both go to infinity, x "wins".

Limit of a Piecewise Function

Problem: f(x) = { x + 1  if x < 2
                { x² - 1 if x ≥ 2

Find lim(x→2) f(x)

Left-hand limit (x→2⁻, use x + 1):
  lim(x→2⁻) (x + 1) = 2 + 1 = 3

Right-hand limit (x→2⁺, use x² - 1):
  lim(x→2⁺) (x² - 1) = 4 - 1 = 3

Both sides equal 3.
Result: lim(x→2) f(x) = 3

Note: f(2) = 2² - 1 = 3, so f is also continuous at x = 2.

Limit Used in Derivative Definition

The derivative of f(x) = x² is defined as a limit:

f'(x) = lim(h→0) [f(x+h) - f(x)] / h
      = lim(h→0) [(x+h)² - x²] / h
      = lim(h→0) [x² + 2xh + h² - x²] / h
      = lim(h→0) [2xh + h²] / h
      = lim(h→0) h(2x + h) / h
      = lim(h→0) (2x + h)
      = 2x

Result: The derivative of x² is 2x.

This shows how limits are the foundation of differential calculus — every derivative is defined as a limit.