Limits of sequences and functions

The limit is the foundation of calculus: both the derivative and the integral are defined through limits. Exams mainly feature limits of sequences, resolving the indeterminate forms 0/0 and ∞/∞, and the special (remarkable) limits. Each type of indeterminate form has its own "key" — let us learn to recognise them.

§1. Limit of a sequence

If, as n grows without bound, the terms aₙ get arbitrarily close to a number A, then A is the limit of the sequence: lim aₙ = A (n → ∞).

Limit laws: the limit of a sum, difference, product and (if the limit of the denominator is not 0) quotient equals the sum, difference, product and quotient of the limits.
Not every sequence has a limit: (−1)ⁿ alternates between −1 and 1 — no limit; n² tends to infinity.

Worked example

Find lim (3n + 1)/(n + 2) (n → ∞).

1. Divide the numerator and denominator by n: (3 + 1/n)/(1 + 2/n).
2. 1/n → 0, 2/n → 0.
3. The limit is 3/1 = 3.

§2. Limit of a function and continuity

If f(x) approaches A as x approaches a, then lim f(x) = A (x → a). The value f(a) itself does not matter — the function may not even be defined at a.
If lim f(x) = f(a), the function f is continuous at a. Polynomials, rational functions (where the denominator is non-zero), roots, exponential and trigonometric functions are continuous on their domains. So the first step is always to substitute the value.

If substitution gives an indeterminate form such as 0/0, ∞/∞, ∞ − ∞, the expression must be transformed first.

One-sided limits. x may approach a only from the left (x → a − 0) or only from the right (x → a + 0). The limit exists only if both one-sided limits are equal. For example, for f(x) = |x|/x we get 1 as x → 0 + 0 and −1 as x → 0 − 0; they differ, so there is no limit as x → 0. Similarly, 1/x has no finite limit as x → 0: it tends to +∞ from the right and to −∞ from the left.

§3. The form 0/0

If both numerator and denominator become 0, they share a factor (x − a). We extract and cancel it:

• Factorisation (for polynomials).
• Multiplying by the conjugate (for roots): multiply √A − B by √A + B.

Cancelling is legitimate because x ≠ a while x → a.

Worked example

Find lim(x→3)(x² − 5x + 6)/(x² − 9).

1. Substitution gives 0/0.
2. x² − 5x + 6 = (x − 2)(x − 3), x² − 9 = (x − 3)(x + 3).
3. Cancel: (x − 2)/(x + 3).
4. At x = 3: 1/6.

Worked example

Find lim(x→0)(√(x + 1) − 1)/x.

1. Multiply by the conjugate: (x + 1 − 1)/(x(√(x + 1) + 1)) = 1/(√(x + 1) + 1).
2. At x = 0: 1/2.

§4. The forms ∞/∞ and ∞ − ∞

In a rational expression, divide the numerator and denominator by the highest power of x. This gives the rule:

• lim(x→∞)(3x² + 1)/(x² − 5) = 3.
• lim(x→∞)(2x + 1)/(x² + 1) = 0.

`∞ − ∞`: turn the expression into a fraction — usually by multiplying by the conjugate.

Worked example

Find lim(n→∞)(√(n² + n) − n).

1. Multiply by the conjugate: (n² + n − n²)/(√(n² + n) + n) = n/(√(n² + n) + n).
2. Divide by n: 1/(√(1 + 1/n) + 1).
3. The limit is 1/(1 + 1) = 1/2.

§5. Special limits

Consequences of the first:

Consequence of the second: lim(n→∞)(1 + k/n)ⁿ = eᵏ.

Worked example

Find lim(x→0) sin 3x/sin 2x.

1. sin 3x/sin 2x = (sin 3x/(3x))·(2x/sin 2x)·(3/2).
2. The first two factors tend to 1.
3. Answer: 3/2.

Worked example

Find lim(n→∞)(1 + 2/n)ⁿ.

1. Let m = n/2: (1 + 1/m)²ᵐ = ((1 + 1/m)ᵐ)².
2. Answer: e².

§6. Applications of limits: asymptotes and the definition of the derivative

A limit is not an end in itself: it describes how a graph behaves ‘at the edges’ and underlies the idea of the derivative.
Asymptotes. A line that the graph approaches without bound is called an asymptote.

For a rational function, look for vertical asymptotes among the zeros of the denominator (if the numerator also vanishes there, cancel first — there may be no asymptote at all). The horizontal asymptote comes from the ∞/∞ rule in §4: if the degrees of numerator and denominator are equal, b is the ratio of the leading coefficients; if the denominator’s degree is larger, b = 0.
Definition of the derivative. The derivative of f(x) at x₀ is the limit of the ratio of increments:

This limit always produces the indeterminate form 0/0, which is resolved by the method of §3 — simplify the numerator and cancel h. That is why a firm grasp of limits is needed to understand the table of derivatives.

Worked example

a) Evaluate lim(h→0) ((2 + h)² − 4)/h. b) Find the asymptotes of y = (3x + 1)/(x − 2).

1. a) Substitution gives 0/0. Expand the numerator: 4 + 4h + h² − 4 = h(4 + h).
2. Cancel h: lim(h→0)(4 + h) = 4. This is the value of (x²)′ at x = 2: 2·2 = 4.
3. b) The denominator vanishes at x = 2 while the numerator equals 7 ≠ 0, so y → ±∞ as x → 2: x = 2 is a vertical asymptote.
4. The degrees are equal: lim(x→∞)(3x + 1)/(x − 2) = 3/1 = 3, so y = 3 is a horizontal asymptote.

Check yourself

1. lim(n→∞)(5n − 2)/(2n + 7) = ?

2. lim(x→1)(x² − 1)/(x − 1) = ?

3. lim(x→0) sin 4x/(2x) = ?

4. lim(x→∞)(x³ − x)/(2x³ + 5) = ?

5. lim(x→4)(√x − 2)/(x − 4) = ?

6. lim(n→∞)(1 + 3/n)ⁿ = ?

7. lim(h→0) ((3 + h)² − 9)/h = ?

8. What is the horizontal asymptote of y = (2x − 1)/(x + 3)?

Common mistakes

• Setting 0/0 equal to 0 or 1 — it is not an answer but an indeterminate form.
• In ∞/∞, comparing only leading coefficients without checking whether the degrees are equal.
• Using the special limit as x → ∞: lim(x→∞) sin x/x = 0, not 1.
• Treating (1 + 1/n)ⁿ as 1ⁿ = 1 — it is the form 1^∞, and the answer is e.
• Forgetting to multiply the denominator too when multiplying by the conjugate.
• Treating every zero of the denominator as a vertical asymptote: if the numerator also vanishes there, cancel first.