Key results:

• lim(x->0) $\frac{e^x - 1}{x}$ = 1
• lim(x->0) $\frac{a^x - 1}{x}$ = ln(a)
• lim(x->0) ln$\frac{1+x}{x}$ = 1
• lim(x->0) (e^(ax) - e^(bx))/x = a - b

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

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