Standard Limits and Computing Function Limits

The standard limits are lim_x→0 sin x / x = 1, lim_x→0 (1 − cos x)/x² = 1/2, lim_x→0 (e^x − 1)/x = 1, lim_x→0 ln(1 + x)/x = 1 and lim_x→∞ (1 + 1/x)^x = e; they are building blocks for computing many other limits. In brief: lim_x→0 sin 3x/x = 3 · lim sin 3x/(3x) = 3, and lim_x→0 (e^2x − 1)/x = 2. The method is to rewrite the expression so that it matches a standard limit.

Knowing a small number of standard limits and a few techniques allows you to compute a large class of limits without l’Hôpital’s rule. This guide lists the standard limits with brief justifications, shows the technique of substitution of the variable, treats limits of the form 1^∞ (via e), gives a systematic approach to computing limits and a table of equivalences (asymptotic approximations near 0).

What are the standard limits?

• lim_x→0 sin x/x = 1; lim_x→0 tan x/x = 1; lim_x→0 arcsin x/x = 1; lim_x→0 arctan x/x = 1.
• lim_x→0 (1 − cos x)/x² = 1/2.
• lim_x→0 (e^x − 1)/x = 1; lim_x→0 (a^x − 1)/x = ln a.
• lim_x→0 ln(1 + x)/x = 1.
• lim_x→0 ((1 + x)^k − 1)/x = k.
• lim_x→∞ (1 + 1/x)^x = e; lim_x→0 (1 + x)^1/x = e.

The first limit follows from the squeeze theorem. The second follows from the first: (1 − cos x)/x² = 2 sin²(x/2)/x² = (1/2)(sin(x/2)/(x/2))² → 1/2. The exponential and logarithm limits express the fact that the derivatives of e^x at 0 and of ln x at 1 equal 1.

How do you compute trigonometric limits?

Rewrite the expression so that the argument of the sine in the numerator equals the denominator (or its multiple).

Example 1

lim_x→0 sin 5x/x = 5 lim sin 5x/(5x) = 5. And lim_x→0 sin 3x/sin 7x = lim [sin 3x/(3x)] · [7x/sin 7x] · (3/7) = 3/7.

Example 2

lim_x→0 tan 4x/(2x) = lim (sin 4x/(4x)) · (1/cos 4x) · 2 = 2. And lim_x→0 (1 − cos 6x)/x² = 36 lim (1 − cos 6x)/(6x)² = 36/2 = 18.

Example 3: a shifted point

lim_x→π sin x/(π − x): let t = π − x → 0; sin x = sin(π − t) = sin t, so the limit is lim sin t/t = 1.

Example 4: using identities

lim_x→0 (cos x − cos 3x)/x² = (1 − cos 3x)/x² − (1 − cos x)/x² → 9/2 − 1/2 = 4.

Example 5: sin(1/x)

lim_x→∞ x sin(1/x) = lim_t→0⁺ sin t/t = 1 with t = 1/x.

How do you compute exponential and logarithmic limits?

Example 6

lim_x→0 (e^3x − 1)/x = 3 lim (e^3x − 1)/(3x) = 3. And lim_x→0 (2^x − 1)/x = ln 2 ≈ 0.6931. And lim_x→0 (e^x − e^−x)/(2x) = lim (1/2)[(e^x − 1)/x + (1 − e^−x)/x] = (1/2)(1 + 1) = 1.

Example 7

lim_x→0 ln(1 + 4x)/x = 4. And lim_x→1 ln x/(x − 1): let t = x − 1 → 0, so ln(1 + t)/t → 1. And lim_x→0 (e^x − 1)/sin x = lim [(e^x − 1)/x] · [x/sin x] = 1.

Example 8: general bases

lim_x→0 (a^x − b^x)/x = lim [(a^x − 1) − (b^x − 1)]/x = ln a − ln b = ln(a/b). For a = 5 and b = 2: ln 2.5 ≈ 0.9163.

How do you handle 1^∞ forms?

For lim f(x)^g(x) with f → 1 and g → ∞, write f = 1 + u with u → 0, so f^g = [(1 + u)^1/u]^ug → e^{lim ug}. In other words: the limit is e^{lim (f − 1)g}.

Example 9

lim_x→∞ (1 + 3/x)^x: (f − 1)g = (3/x) · x = 3, so the limit is e³ ≈ 20.09.

Example 10

lim_x→∞ ((x + 2)/(x − 1))^2x: f − 1 = 3/(x − 1), so (f − 1)g = 6x/(x − 1) → 6, and the limit is e⁶.

Example 11

lim_x→0 (cos x)^1/x2: f − 1 = cos x − 1 ≈ −x²/2, so (f − 1)g → −1/2 and the limit is e^−1/2 ≈ 0.6065.

What is a systematic approach?

1. Try direct substitution. If you get a number, you are done (for continuous functions).
2. If you get 0/0, ∞/∞ or another indeterminate form, simplify: factor, cancel, use conjugates, common denominators.
3. Match with a standard limit after a substitution t = g(x).
4. For 1^∞, use the exponent rule above.
5. If all else fails, use l’Hôpital’s rule or series expansions.
6. Check the result numerically with a value of x close to the limit point.

Table of equivalent functions near 0

Two functions are equivalent near 0 (written ∼) if their ratio tends to 1. In limits of products and quotients, an expression may be replaced by an equivalent one.

Equivalents can be substituted in factors, but not blindly in sums: sin x − x is not equivalent to 0; its leading behaviour is −x³/6.

Where do these limits come from?

The limit sin x/x was used implicitly by Archimedes in his comparison of the circumference and the polygons and by Newton and Euler in finding the derivative of sine. The number e as a limit of (1 + 1/n)ⁿ was found by Jacob Bernoulli in 1683 when studying compound interest. The systematic use of these limits to derive the derivatives of elementary functions, as, is part of standard calculus since the nineteenth century.

Where are they used?

The small-angle approximations sin x ≈ x and cos x ≈ 1 − x²/2 are used in physics (the period of a pendulum for small swings), in optics and in engineering. The limit (1 + r/n)ⁿ → e^r gives continuous compounding in finance. The approximations ln(1 + x) ≈ x and e^x ≈ 1 + x are used for estimates of small percentage changes.

Which mistakes are common?

Applying sin x/x → 1 when x does not tend to 0

The limit holds only when the argument tends to 0. For x → ∞, sin x/x → 0.

Not matching the argument and the denominator

sin 3x/x → 3, not 1. Multiply and divide by 3.

Replacing terms in a sum by equivalents

(sin x − x)/x³: replacing sin x by x gives 0/x³ = 0, which is wrong. The true limit is −1/6.

Forgetting the e in the 1^∞ form

(1 + 1/n)ⁿ does not tend to 1. The limit is e.