Key results:

• 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) (e^(ax) - e^(bx))/x = a - b

Useful manipulation: For lim(x->0) (a^x - b^x)/x: = lim(x->0) [(a^x - 1) - (b^x - 1)]/x = lim(x->0) (a^x - 1)/x - (b^x - 1)/x = ln(a) - ln(b) = ln(a/b)

For expressions like lim(x->0) (e^(sin x) - 1)/x: Rewrite as [(e^(sin x) - 1)/sin x] * [sin x / x] = 1 * 1 = 1 This "chain substitution" technique is powerful: if f(x) -> 0, then (e^(f(x)) - 1)/f(x) -> 1.

Logarithmic limits at infinity:

• lim(x->infinity) ln(x)/x = 0 (log grows slower than linear)
• lim(x->0+) x * ln(x) = 0 (substitute x = e^(-t), t -> infinity, get -t/e^t -> 0)