Squeeze (Sandwich, or Pinching) Theorem. Let f, g, h be three functions defined on an
interval [a, b]. If a < c < b and f (x) ≤ g(x) ≤ h(x) for all a ≤ x ≤ b and limx→c f (x) = L, and
limx→c h(x) = L, then
lim g(x) = L
x→c
Proof. It is required to show that for any > 0 there is a δ > 0 such that
(∗) 0 < |x − c| < δ ⇒ |g(x) − L| < So let > 0. We are given limx→c f (x) = L, and limx→c h(x) = L. Therefore we are given
(i) there is δ1 > 0 such that 0 < |x − c| < δ ⇒ |f (x) − L| < and
(ii) there is δ2 > 0 such that 0 < |x − c| < δ ⇒ |h(x) − L| < We are also given
(iii)
f (x) ≤ g(x) ≤ h(x) for all a ≤ x ≤ b.
Given > 0 we must find δ > 0 so that (*) holds: We take δ = min{δ1 , δ2 }, and verify (*):
Let 0 < |x − c| < δ, since δ is the smaller of δ1 , δ2 , both (i) and (ii) hold, hence |f (x) − L| < and |h(x) − L| < . These are equivalent to
L − < f (x) < L + and L − < h(x) < L + . Thus
L − ≤ f (x) ≤ g(x) ≤ h(x) ≤ L + which is the same as saying
|g(x) − L| < .
Exercise: Application of the Squeeze Theorem. Show that the following function g1 is
NOT continuous at x = 0
x sin x1 if x 6= 0
g1 (x) =
0
if x = 0
[HINT: Find a sequence (xn )n such that limn→∞ xn = 0 and g1 (xn ) = 1 for all n = 1, 2, 3 · · ·].
On the other hand, by using the Squeeze Theorem, show that the following function g2 is continuous at 0
2
x sin x1 if x 6= 0
g2 (x) =
0
if x = 0
[HINT: It suffices to assume the function g2 (x) is defined over the interval [−1, 1]. To apply the
squeeze theorem to g2 , take a = −1, b = 1, c = 0, L = 0, and select appropriate squeezing functions.
Graph of g2