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