2.3 Suppose c ∈ R and suppose lim_{x→a}f(x) and lim_{x→a}g(x) exist. Then lim_{x→a}[f(x) + g(x)] = lim_{x→a}f(x) + lim_{x→a}g(x) lim_{x→a}[cf(x)] = c lim_{x→a}f(x) lim_{x→a}[f(x)g(x)] = lim_{x→a}f(x) lim_{x→a}g(x) lim_{x→a} \frac{f(x)}{g(x)} = \frac{lim_{x→a}f(x)}{lim_{x→a}g(x)} if lim_{x→a}g(x) ≠ 0 Defn: f is continuous at a iff lim_{x→a}f(x) = f(lim_{x→a}x) = If f is continuous then

lim_{x→a}f(g(x)) = f(lim_{x→a}g(x))

Theorem: If f(x) ≤ g(x) near a (except possibly at a) and if lim_{x→a}f(x) and lim_{x→a}g(x) exist, then

lim_{x→a}f(x) ≤ lim_{x→a}g(x)

Squeeze theorem: If f(x) ≤ g(x) ≤ h(x) near a (except possibly at a) and if lim_{x→a}f(x) = L and lim_{x→a}h(x) = L, then

lim_{x→a}g(x) = L

Example: g(x) = xsin\frac{1}{x} Defn: lim_{x→a}f(x) = L if x close to a (except possibly at a) implies f(x) is close to L.

Defn: lim_{x→a}f(x) = L if x close to a (except possibly at a) implies f(x) is close to L.

Defn: lim_{x→a}f(x) = L if for all ϵ > 0, there exists a δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ϵ

Show lim_{x→1}2 = Show lim_{x→4}2x + 3 = Defn: lim_{x→a^{−}}f(x) = L if x close to a and x < a implies f(x) is close to L.

Defn: lim_{x→a^{+}}f(x) = L if x close to a and x > a implies f(x) is close to L.

Defn: lim_{x→a}f(x) = ∞ if x close to a (except possibly at a) implies f(x) is large.

Defn: lim_{x→a}f(x) = −∞ if x close to a (except possibly at a) implies f(x) is negative and |f(x)| is large.