Limits of sequences and functions — Mathematics

In this lesson you will learn

Find limits of sequences: divide by the highest power of n, multiply by the conjugate and use (1 + k/n)ⁿ → eᵏ
Compute limits of functions at a point, one-sided limits and limits at infinity, including rational functions at ∞ by comparing degrees (also with a parameter)
Resolve 0/0 by factoring, by the conjugate and with the standard limits, including trigonometric limits at points such as π/4
Check continuity at a point and find a parameter from the continuity condition

Murad stands 1 metre from a door, and with every step he covers half of the remaining distance. After the first step he has walked 1/2 m, after the second 3/4 m, after the third 7/8 m, and after n steps 1 − 1/2ⁿ metres. These numbers — 0.5, 0.75, 0.875, 0.9375, … — never equal 1, yet they get as close to 1 as we like. Mathematicians say it briefly: the limit of this sequence is 1. The limit is the foundation of calculus: in the next lessons we build both the derivative and the integral from it. In stage II of the DİM entrance exam this topic appears as «Funksiyanın limiti» (the limit of a function); in 2026 the exams of both group I and group II contained a coded limit task — a task whose answer is a single number.

The limit of a sequence

From the lesson “Sequences and arithmetic progressions” we know that a sequence is a numbered list of numbers: a₁, a₂, a₃, …, aₙ, … Now we are interested not in single terms but in where the terms go as n grows. For example, the terms of aₙ = 1/n are 1, 0.5, 0.33…, 0.25, …, 0.01 (n = 100), 0.001 (n = 1000), and they approach zero. The sequence aₙ = (−1)ⁿ, on the other hand, swings between −1 and 1 and approaches no number at all.

Definition

Limit of a sequence

A number A is the limit of the sequence (aₙ) if for every small positive number ε there is an index N such that |aₙ − A| < ε for all n > N. In plain words: from some index on, all terms lie in any narrow band around A that we choose. Notation: lim (n→∞) aₙ = A. A sequence that has a limit is convergent; one that has no limit is divergent.

The definition at work

1) Show that the limit of aₙ = (2n + 1)/n is 2, and find N for ε = 0.01.
2) Do the sequences aₙ = (−1)ⁿ and bₙ = n² have limits?

Show solution

1) aₙ = 2 + 1/n, so |aₙ − 2| = 1/n.
1/n < 0.01 ⇔ n > 100. So N = 100: from the 101st term on, all terms lie in the interval (1.99, 2.01). However small ε is, we can take N = 1/ε, so lim (n→∞) aₙ = 2.
2) (−1)ⁿ: the terms are −1, 1, −1, 1, … They stay neither near 1 nor near −1, so there is no limit — the sequence diverges.
n²: the terms 1, 4, 9, 16, … grow without bound. There is no finite limit; people write lim (n→∞) n² = ∞, but ∞ is not a number, and the sequence still diverges.

lim (n→∞) c = c lim (n→∞) 1/n = 0 lim (n→∞) 1/nᵏ = 0 (k > 0) lim (n→∞) qⁿ = 0 (|q| < 1)lim (n→∞) c = c lim (n→∞) 1/n = 0 lim (n→∞) 1/nᵏ = 0 (k > 0) lim (n→∞) qⁿ = 0 (|q| < 1)

where:

ca constant
ka positive exponent
qa number with |q| < 1 (like the ratio of an infinite decreasing geometric progression)

Basic limits: we always reduce complicated expressions to these simple ones.

lim (aₙ ± bₙ) = A ± B lim (aₙ · bₙ) = A · B lim (aₙ/bₙ) = A / B (B ≠ 0)lim (aₙ ± bₙ) = A ± B lim (aₙ · bₙ) = A · B lim (aₙ/bₙ) = A / B (B ≠ 0)

where:

A, Blim aₙ = A and lim bₙ = B — finite limits

The limit laws apply only when both limits are finite (and, for a quotient, B ≠ 0).

In a sequence given as a fraction both the numerator and the denominator tend to infinity — this is the form ∞/∞, and the laws cannot be applied directly. The main method: divide the numerator and the denominator by the highest power of n in the denominator. Then terms like 1/n and 5/n² tend to zero, and only the coefficients of the leading terms remain.

Sequences given as fractions

Find the limit:
1) aₙ = (3n² − n)/(2n² + 5);
2) aₙ = (4n + 7)/(n² + 1);
3) aₙ = (5 · 3ⁿ + 2ⁿ)/(3ⁿ⁺¹ − 4);
4) aₙ = √(n² + 6n) − n.

Show solution

1) Divide the numerator and the denominator by n²: aₙ = (3 − 1/n)/(2 + 5/n²). As n → ∞, 1/n → 0 and 5/n² → 0, so the limit is 3/2 = 1.5.
2) Again divide by n²: (4/n + 7/n²)/(1 + 1/n²) → 0/1 = 0. The degree of the numerator is less than that of the denominator — the limit is 0.
3) Here the “strongest” term is 3ⁿ. Divide the numerator and the denominator by 3ⁿ: (5 + (2/3)ⁿ)/(3 − 4/3ⁿ). (2/3)ⁿ → 0 (|q| < 1) and 4/3ⁿ → 0, so the limit is 5/3.
4) This is the form ∞ − ∞. Multiply and divide by the conjugate: (n² + 6n − n²)/(√(n² + 6n) + n) = 6n/(√(n² + 6n) + n). Divide the numerator and the denominator by n: 6/(√(1 + 6/n) + 1) → 6/(1 + 1) = 3.

Monotone sequences and the number e

Definition

Monotone and bounded sequence

A sequence (aₙ) is increasing if aₙ₊₁ > aₙ for every n and decreasing if aₙ₊₁ < aₙ; increasing and decreasing sequences are called monotone. A sequence is bounded if there is a number M with |aₙ| ≤ M for all n; if only aₙ ≤ M holds, it is bounded above.

Weierstrass theorem: a monotone bounded sequence has a limit. The reason is easy to see: an increasing sequence climbs with every step but cannot pass the “ceiling” M, so its terms crowd near some number not greater than M. Both conditions matter: (−1)ⁿ is bounded but not monotone; n² is monotone but not bounded — neither has a limit.

Monotonicity and boundedness

1) Show that aₙ = n/(n + 1) is increasing and bounded, and find its limit.
2) Answer the same questions for bₙ = 2 + 5/n.

Show solution

1) Look at the difference: aₙ₊₁ − aₙ = (n + 1)/(n + 2) − n/(n + 1) = ((n + 1)² − n(n + 2))/((n + 1)(n + 2)) = 1/((n + 1)(n + 2)) > 0 — the sequence is increasing.
Since n < n + 1, 0 < aₙ < 1 — the sequence is bounded. By the theorem the limit exists: dividing by n, aₙ = 1/(1 + 1/n) → 1.
2) 5/(n + 1) < 5/n, so bₙ₊₁ < bₙ — the sequence is decreasing; 2 < bₙ ≤ 7 — it is bounded. Limit: 2 + 0 = 2. The terms 7, 4.5, 3.67, … come down to 2 from above.

Now let us look at the most famous sequence: aₙ = (1 + 1/n)ⁿ. The bracket tends to 1 while the exponent tends to infinity. The idea “any power of 1 is 1” is wrong here: the base is not equal to 1, it is slightly larger, and the exponent keeps growing. Let us compute:

n	1	2	5	10	100	1000	10⁶
(1 + 1/n)ⁿ	2	2.25	2.48832	2.59374	2.70481	2.71692	2.71828

The terms increase but never exceed 3 (this can be proved).

lim (n→∞) (1 + 1/n)ⁿ = e ≈ 2.71828lim (n→∞) (1 + 1/n)ⁿ = e ≈ 2.71828

where:

ean irrational number, the base of the natural logarithm (ln x = logₑ x) and of the function eˣ

The second standard limit: (1 + 1/n)ⁿ is increasing and bounded, so it has a limit; that limit is the number e.

lim (n→∞) (1 + k/n)ⁿ = eᵏlim (n→∞) (1 + k/n)ⁿ = eᵏ

where:

ka constant (it may be negative)

Consequence: putting n = k · m gives (1 + 1/m)^(km) = ((1 + 1/m)ᵐ)ᵏ → eᵏ.

Limits with the number e

Find the limit:
1) lim (n→∞) (1 + 1/n)³ⁿ;
2) lim (n→∞) (1 + 2/n)ⁿ;
3) lim (n→∞) ((n − 1)/n)ⁿ;
4) lim (n→∞) (1 + 1/(2n))ⁿ.

Show solution

1) (1 + 1/n)³ⁿ = ((1 + 1/n)ⁿ)³ → e³ ≈ 20.09.
2) k = 2: (1 + 2/n)ⁿ → e² ≈ 7.39.
3) (n − 1)/n = 1 − 1/n, so k = −1: (1 − 1/n)ⁿ → e⁻¹ = 1/e ≈ 0.368.
4) 1/(2n) = (1/2)/n, so k = 1/2: the limit is e^(1/2) = √e ≈ 1.65.

The limit of a function: at a point and at infinity

Now the argument is not an index but a continuous variable x. The function f(x) = (x² − 1)/(x − 1) is not defined at x = 1 (the denominator is zero). But let us look at its values near 1:

x	0.9	0.99	1.01	1.1
f(x)	1.9	1.99	2.01	2.1

f is undefined at x = 1, but the values approach 2: for x ≠ 1, f(x) = x + 1.

Definition

Limit of a function at a point

If the values f(x) get as close as we like to a number A when x approaches a (x ≠ a), then A is the limit of f at the point a: lim (x→a) f(x) = A. The limit depends on the values of the function near a, not on its value at a. If x approaches a only from the left (x < a), we get the left-hand limit; only from the right (x > a), the right-hand limit.

lim (x→a) f(x) = A ⇔ lim (x→a⁻) f(x) = lim (x→a⁺) f(x) = A

where:

lim (x→a⁻)left-hand limit (x < a)
lim (x→a⁺)right-hand limit (x > a)

The limit exists only when both one-sided limits exist and are equal.

One-sided limits

1) f(x) = x + 1 if x < 2; f(x) = 5 − x if x ≥ 2. Does lim (x→2) f(x) exist?
2) Does g(x) = |x|/x have a limit at x = 0?

Show solution

1) Left-hand limit (x < 2, first formula): x + 1 → 2 + 1 = 3. Right-hand limit (x > 2, second formula): 5 − x → 5 − 2 = 3. The one-sided limits are equal, so lim (x→2) f(x) = 3.
2) For x > 0, |x| = x and g(x) = 1; for x < 0, |x| = −x and g(x) = −1. The right-hand limit is 1, the left-hand limit is −1. They differ — there is no limit (the graph has a “jump”).

Polynomials, roots, eˣ, ln x, sin x, cos x and fractions built from them are continuous at every point of their domains (we make this precise in the last section). At such a point the limit equals the value of the function: we simply substitute x = a. Exam limits are “hard” precisely because substitution gives 0/0 or ∞/∞.

By substitution

1) lim (x→2) (x³ − 3x + 1);
2) lim (x→π/3) (2cos x + tan²x);
3) lim (x→1) (x + 3)/(x² + 1).

Show solution

1) 2³ − 3 · 2 + 1 = 8 − 6 + 1 = 3.
2) cos(π/3) = 1/2, tan(π/3) = √3: 2 · 1/2 + (√3)² = 1 + 3 = 4.
3) (1 + 3)/(1 + 1) = 4/2 = 2. The denominator is not zero, so substitution is enough.

Limits at infinity. lim (x→∞) f(x) = A means that f(x) approaches A as |x| grows without bound. The key fact is the same as for sequences: lim (x→∞) c/xᵏ = 0 (k > 0). For a rational function we divide the numerator and the denominator by the highest power of x in the denominator; in the end only the leading terms “survive”.

lim (x→∞) (aₘxᵐ + … + a₀)/(bₖxᵏ + … + b₀) = aₘ/bₖ if m = klim (x→∞) (aₘxᵐ + … + a₀)/(bₖxᵏ + … + b₀) = aₘ/bₖ if m = k

where:

m, kthe degrees of the numerator and the denominator
aₘ, bₖthe leading coefficients (coefficients of the highest-degree terms, ≠ 0)

Limit of a rational function as x → ∞: 0 if m < k, aₘ/bₖ if m = k, no finite limit (∞) if m > k.

Comparing degrees

Find the limit as x → ∞:
1) (6x³ − 2x + 1)/(3x³ + x²);
2) (5x² + 4)/(x³ − 2);
3) (x³ + 1)/(4x² − x);
4) (2x − 1)²/(x² + 5);
5) √(x² + 8x) − x.

Show solution

1) m = k = 3: the limit is 6/3 = 2.
2) m = 2 < k = 3: the limit is 0.
3) m = 3 > k = 2: no finite limit, the expression tends to infinity.
4) Careful: (2x − 1)² = 4x² − 4x + 1, so the leading coefficient is 4, not 2. m = k = 2: the limit is 4/1 = 4.
5) ∞ − ∞ (x > 0). With the conjugate: (x² + 8x − x²)/(√(x² + 8x) + x) = 8x/(√(x² + 8x) + x). Divide by x: 8/(√(1 + 8/x) + 1) → 8/2 = 4.

DİM style: coded task (parameter)

For which value of a is lim (x→∞) ((a² − 4)x³ + (a + 1)x² − 5)/(x² + 3) = 3 true?

Show solution

1) The degree of the denominator is 2. If the x³ term stayed in the numerator, its degree would be 3 > 2 and there would be no finite limit. So a² − 4 = 0, i.e. a = 2 or a = −2.
2) Then the numerator has degree 2 and the limit is the ratio of the leading coefficients: (a + 1)/1 = a + 1.
3) a + 1 = 3 ⇒ a = 2. Check: for a = −2 the limit is −2 + 1 = −1 ≠ 3, so that value does not fit.
Answer: 2.

Interactive

Loading simulation…

The first graph is ((a² − 4)x³ + (a + 1)x² − 5)/(x² + 3), the second is the line y = 3. For a = 2 the graph hugs the line y = 3 on both sides. For a = −2 it approaches the line y = −1, and for other values of a the graph “runs away” up or down — there is no finite limit.

DİM style: closed task

If lim (x→∞) (ax² − 3x + 1)/(4x² + x) = −2, find a.
A) −2 B) −8 C) 8 D) −1/2 E) 2

Show solution

The numerator and the denominator have the same degree (2), so the limit is the ratio of the leading coefficients: a/4 = −2 ⇒ a = −8.
Answer: B. (Option −1/2 catches those who divide −2 by 4, and 8 those who lose the sign.)

The form 0/0 and the standard limits

If substituting x = a gives 0/0, this is not an answer but an indeterminate form: as x → 0, x/x → 1, x²/x → 0, 5x/x → 5 — all of them look like “0/0”, yet the limits differ. So the expression has to be transformed. The key idea: in a limit x ≠ a, so we may cancel the common factor (x − a) of the numerator and the denominator.

1
1. Substitute
Put x = a. If you get a number, that is the answer. If you get c/0 (c ≠ 0), there is no finite limit. If you get 0/0, go to step 2.
2
2. Bring out the factor (x − a)
Factor the polynomials (a is their root). If there is a root sign, multiply the numerator and the denominator by the conjugate. In trigonometry, use identities.
3
3. Cancel and substitute again
Cancel (x − a) (allowed because x ≠ a) and put x = a into the simplified expression.

Factoring

1) lim (x→−2) (x² + 5x + 6)/(x² − 4);
2) lim (x→1) (x³ − 1)/(x² − 1);
3) lim (x→3) (2x² − 5x − 3)/(x² − 9).

Show solution

1) x = −2: (4 − 10 + 6)/(4 − 4) = 0/0.
(x + 2)(x + 3)/((x + 2)(x − 2)) = (x + 3)/(x − 2) → 1/(−4) = −0.25.
2) 0/0. Difference of cubes: x³ − 1 = (x − 1)(x² + x + 1); x² − 1 = (x − 1)(x + 1).
(x² + x + 1)/(x + 1) → 3/2 = 1.5.
3) 0/0. The roots of 2x² − 5x − 3 are 3 and −1/2, so 2x² − 5x − 3 = (x − 3)(2x + 1).
(2x + 1)/(x + 3) → 7/6.

The conjugate

1) lim (x→4) (√x − 2)/(x − 4);
2) lim (x→0) (√(9 + x) − 3)/x;
3) lim (x→5) (x − 5)/(√(x + 4) − 3);
4) Coded task: compute lim (x→1) (√(x + 3) − 2)/(x² − 1).

Show solution

1) x − 4 = (√x − 2)(√x + 2) (difference of squares), so the expression becomes 1/(√x + 2) → 1/(2 + 2) = 1/4.
2) Multiply the numerator and the denominator by √(9 + x) + 3: (9 + x − 9)/(x(√(9 + x) + 3)) = 1/(√(9 + x) + 3) → 1/6.
3) The conjugate is in the denominator: (x − 5)(√(x + 4) + 3)/(x + 4 − 9) = √(x + 4) + 3 → 3 + 3 = 6.
4) 0/0. Multiply and divide by √(x + 3) + 2: (x + 3 − 4)/((x² − 1)(√(x + 3) + 2)) = (x − 1)/((x − 1)(x + 1)(√(x + 3) + 2)) = 1/((x + 1)(√(x + 3) + 2)) → 1/(2 · 4) = 1/8.
Answer: 0.125.

Now to the trigonometric 0/0. When the angle is measured in radians, sin x is almost equal to x for small x: on the unit circle a small arc hardly differs from its chord. In numbers: sin 0.5 / 0.5 ≈ 0.9589; sin 0.1 / 0.1 ≈ 0.99833; sin 0.01 / 0.01 ≈ 0.99998. The ratio approaches 1.

lim (x→0) sin x / x = 1lim (x→0) sin x / x = 1

where:

xan angle in radians

The first standard (fundamental trigonometric) limit.

lim (x→0) sin kx / x = k lim (x→0) tan x / x = 1 lim (x→0) (1 − cos x)/x² = 1/2lim (x→0) sin kx / x = k lim (x→0) tan x / x = 1 lim (x→0) (1 − cos x)/x² = 1/2

where:

ka constant, k ≠ 0

Consequences: sin kx / x = k · sin kx / (kx); tan x / x = (sin x / x) · (1 / cos x); 1 − cos x = 2sin²(x/2) ≈ 2 · (x/2)² = x²/2.

Trigonometric limits as x → 0

1) lim (x→0) sin 7x / (2x);
2) lim (x→0) tan 3x / sin 5x;
3) lim (x→0) (1 − cos 6x)/x²;
4) lim (x→0) (1 − cos 2x)/(x · sin 3x).

Show solution

1) sin 7x / (2x) = (7/2) · sin 7x / (7x) → 7/2 = 3.5.
2) tan 3x / sin 5x = (tan 3x / (3x)) · (5x / sin 5x) · (3x/(5x)) → 1 · 1 · 3/5 = 0.6.
3) 1 − cos 6x = 2sin²3x, so (1 − cos 6x)/x² = 2 · (sin 3x / x)² → 2 · 3² = 18.
4) 1 − cos 2x = 2sin²x. The expression is 2sin²x / (x · sin 3x) = 2 · (sin x / x) · (sin x / sin 3x). Since sin x / sin 3x → 1/3, the limit is 2 · 1 · 1/3 = 2/3.

Interactive

Loading simulation…

y = sin kx / x is not defined at x = 0, but as x → 0 the graph approaches the line y = k — sin kx / x → k. Change k and check. As x grows, the graph dies down to zero: sin kx / x → 0 as x → ∞.

When x → a ≠ 0. For example, as x → π/4 the formula sin x / x does not apply directly. If substitution gives 0/0 there are two ways: 1) use identities to bring out the factor that becomes zero and cancel it: cos 2x = (cos x − sin x)(cos x + sin x), 4sin²x − 1 = (2sin x − 1)(2sin x + 1), sin 2α = 2sin α cos α; 2) substitute t = x − a (t → 0) and use the reduction formulas.

Trigonometric limits at a point other than 0

1) lim (x→π/4) cos 2x / (cos x − sin x);
2) lim (x→π/6) (4sin²x − 1)/(2sin x − 1);
3) lim (x→π/3) sin 3x / sin 6x;
4) lim (x→π) sin x / (x − π).

Show solution

1) cos(π/2) = 0 and cos(π/4) − sin(π/4) = 0 — 0/0.
cos 2x = (cos x − sin x)(cos x + sin x); after cancelling, cos x + sin x → √2/2 + √2/2 = √2 ≈ 1.41.
2) sin(π/6) = 1/2: 0/0. Difference of squares: (2sin x − 1)(2sin x + 1)/(2sin x − 1) = 2sin x + 1 → 2 · 1/2 + 1 = 2.
3) sin π = 0, sin 2π = 0 — 0/0. sin 6x = 2sin 3x cos 3x, so the expression is 1/(2cos 3x) → 1/(2cos π) = −1/2.
4) t = x − π → 0. sin x = sin(π + t) = −sin t, so the expression is −sin t / t → −1.

DİM style: coded task

Compute lim (x→π/4) (1 − tan x)/(cos 2x).

Show solution

1) tan(π/4) = 1, cos(π/2) = 0 — 0/0.
2) 1 − tan x = (cos x − sin x)/(cos x) and cos 2x = (cos x − sin x)(cos x + sin x).
3) After cancelling: 1/(cos x · (cos x + sin x)).
4) x = π/4: 1/((√2/2) · √2) = 1/1 = 1.
Answer: 1.

Increments and continuity

The increment of the argument Δx = x − x₀ and the increment of the function Δf = f(x₀ + Δx) − f(x₀) are the language of the next lesson, “Increments and the definition of the derivative”. In this language continuity reads: if Δf → 0 as Δx → 0, then f is continuous at x₀. That is, a small change of the argument causes only a small change of the function, and the graph has no break.

The increments shrink

f(x) = x² − 3x, x₀ = 2. Compute Δf for Δx = 0.1 and Δx = 0.01, then express Δf in terms of Δx.

Show solution

f(2) = 4 − 6 = −2.
Δx = 0.1: f(2.1) = 4.41 − 6.3 = −1.89; Δf = −1.89 − (−2) = 0.11.
Δx = 0.01: f(2.01) = 4.0401 − 6.03 = −1.9899; Δf = 0.0101.
In general: f(2 + Δx) = 4 + 4Δx + (Δx)² − 6 − 3Δx = −2 + Δx + (Δx)², so Δf = Δx + (Δx)² → 0. The function is continuous at x₀ = 2.

Definition

Continuity at a point

A function f is continuous at x₀ if three conditions hold: 1) f(x₀) is defined; 2) lim (x→x₀) f(x) exists (the left-hand and right-hand limits are equal); 3) this limit equals f(x₀). If one of the conditions fails, x₀ is a point of discontinuity.

lim (x→x₀) f(x) = f(x₀) ⇔ lim (Δx→0) Δf = 0

where:

x₀the point under study
Δx, Δfthe increments of the argument and of the function

Two equivalent forms of continuity: the limit equals the value, or a small increment gives a small increment.

1
1. Left-hand limit
Let x → x₀ in the formula written for x < x₀.
2
2. Right-hand limit
Let x → x₀ in the formula written for x > x₀.
3
3. The value at the point
Compute f(x₀) with the formula given for x = x₀.
4
4. Equate and check
Set the three numbers equal, find the parameter and check the value you found.

Continuity and a parameter

For which value of a is the function continuous?
1) f(x) = (x² − 4)/(x − 2), x ≠ 2; f(2) = a.
2) f(x) = x² + a if x < 1; f(x) = 3x − 1 if x ≥ 1.
3) f(x) = sin 3x / x, x ≠ 0; f(0) = a + 1.

Show solution

1) lim (x→2) (x − 2)(x + 2)/(x − 2) = lim (x→2) (x + 2) = 4. For continuity f(2) = 4, i.e. a = 4.
2) Left-hand limit: 1 + a. Right-hand limit and f(1): 3 · 1 − 1 = 2. 1 + a = 2 ⇒ a = 1.
3) lim (x→0) sin 3x / x = 3, so a + 1 = 3 ⇒ a = 2.

DİM style: a fully written solution

f(x) = ax − 1 if x < 1; f(1) = 3; f(x) = x² + b if x > 1. If the function is continuous at x = 1, find a + b.

Show solution

Solution. By the definition of continuity, the left-hand limit, the right-hand limit and f(1) must be equal.
1) Left-hand limit: for x < 1, f(x) = ax − 1, so lim (x→1⁻) f(x) = a − 1.
2) Right-hand limit: for x > 1, f(x) = x² + b, so lim (x→1⁺) f(x) = 1 + b.
3) By the condition, f(1) = 3.
4) a − 1 = 3 ⇒ a = 4; 1 + b = 3 ⇒ b = 2.
5) Check: for a = 4, b = 2 both one-sided limits equal 3 and f(1) = 3 — the function is continuous.
Answer: a + b = 6.

Coded tasks: write the answer as a number

1.lim (n→∞) (5n² + 1)/(2n² − n) =
2.lim (x→3) (x² − 9)/(x − 3) =
3.lim (x→0) sin 6x / (3x) =
4.lim (x→0) (1 − cos 4x)/x² =
5.lim (x→∞) (√(x² + 10x) − x) =
6.lim (n→∞) (1 + 3/n)ⁿ = eᵏ, k =

Key points

lim (n→∞) aₙ = A: from some index on, all terms lie in any narrow band around A; for a sequence given as a fraction, divide the numerator and the denominator by the highest power of n.
A monotone bounded sequence has a limit (Weierstrass theorem); (1 + 1/n)ⁿ → e ≈ 2.718 and (1 + k/n)ⁿ → eᵏ.
lim (x→a) f(x) exists only when the left-hand and right-hand limits are equal; as x → ∞ the degrees decide the limit of a rational function: 0 if m < k, the ratio of the leading coefficients if m = k, no finite limit if m > k.
0/0 is not an answer: factor, multiply by the conjugate or use the standard limits — sin x / x → 1, sin kx / x → k, (1 − cos x)/x² → 1/2 (x → 0, radians).
When x → a ≠ 0, a trigonometric 0/0 is resolved with identities such as cos 2x = (cos x − sin x)(cos x + sin x), or with the substitution t = x − a.
f is continuous at x₀ ⇔ lim (x→x₀) f(x) = f(x₀) ⇔ Δf → 0 as Δx → 0; in a parameter problem, set the left-hand limit, the right-hand limit and f(x₀) equal.

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What is lim (n→∞) (1 + 1/n)ⁿ?