Calculating limits and equivalents using DLs

Why use limit expansions for limits?

Limit expansions allow us to resolve indeterminate forms (0/0, ∞/∞, 1^∞, etc.) by replacing a complex function with an equivalent polynomial in the neighbourhood of the point under consideration.

Method

1. Identify the point to which x tends (often 0, sometimes +∞ via x = 1/t).
2. Choose a degree of the limit series sufficient to ensure that the first non-zero term of each expression appears.
3. Replace each term with its series expansion, simplify, then take the limit.

Example: Calculate lim x→0 of (e^x - 1 - x) / x²

The second-order power series expansion of e^x is: e^x = 1 + x + x²/2 + o(x²). Therefore, e^x - 1 - x = x²/2 + o(x²). The quotient becomes: (x²/2 + o(x²)) / x² = 1/2 + o(1) → 1/2

Equivalents from a power series

The first non-zero term of a DL provides an equivalent: if f(x) = a_k(x – a)^k + o((x – a)^k) with a_k ≠ 0, then f(x) ~ a_k(x – a)^k in the neighbourhood of a.

Example: sin(x) – x ~ –x³/6 at 0, because sin(x) = x – x³/6 + o(x³).

Common pitfall

Never use an equivalent in a sum or difference if this amounts to subtracting two terms of the same order: useful information is lost. You must therefore raise the term to the next higher power, as in the example e^x – 1 – x above. Equivalents can be handled easily for a product or a quotient, but not for a sum or a difference.