Theorem 1:
Numerical: Evaluate Lim xà 1 [( x 15-1)/( x 10-1)] Solution: Rewrite the expression as Lim xà 1 [( x 15-1)/( x-1)] * Lim xà 1 [( x-1)/( x 10-1)] = 15( 1) 14 / 10( 1) 9 = 3 / 2
Theorem 2:
Let f and g be two real valued functions with the same domain such that f( x) ≤ g( x) for all x in the domain of definition, For some a, if both lim x→a f( x) and lim x→a g( x) exist, then lim x→a f( x) ≤ lim x→a g( x).
Sandwich Theorem 3:
Let f, g and h be real functions such that f( x) ≤ g( x) ≤ h( x) for all x in the common domain of definition. For some real number a, if lim x→a f( x) = l = lim x→a h( x), then lim x→a g( x) = l.
Theorem 4: lim x→0( sin x)/ x = 1
Theorem 5: lim x→0( 1- cos x)/ x = 0
Numerical: Evaluate lim x→0( sin 5x)/( Sin 4x)
Solution: We can rewrite the expression as lim x→0( sin 5x)/ x) * lim x→0 x /( Sin 4x)
= 5 / 4 * lim x→0( sin 5x)/ 5x) / lim x→0( Sin 4x)/ 4 = 5 / 4 * 1 * 1 = 5 / 4