
Limits and Continuity 3-21
x
x
x
x
x
x
2
0
2
1
0 0
1
0
sin
lim sin .
→ →
=
→
as
i.e.,
The following theorem generalises the discussion that we so far had.
THEOREM 3.5 [The “Sandwich” Theorem] Assume that g(x) £ f(x) £ h(x) for any x in an interval
around the point a. If
lim ( )
x a
h x L
→
= and lim ( ) ,
x a
g x L
→
= then lim ( ) .
x a
f x L
→
=
Proof: First we prove the special case that g(x) = 0 for all x.
So suppose g(x) = 0 for all x, then L = 0. Also, lim ( ) .
x a
h x
→
= 0 Now, we need to prove lim ( ) .
x a
f x
→
= 0
By the definition of the limit lim ( ) ,
x a
h x
→
= 0 we have for e > 0 there exists d > 0 such that if
0 < ½x - a½ < d, then ½