Calculus I: Sandwich theorem

1. Statement

The squeeze theorem, known also as the sandwich theorem or Pinching theorem is used to evaluate the limit of a function by comparison with two other functions whose limits are known or easily calculated.

Let I be an interval having the point "a".

Let f, g, and h be functions defined on I, except possibly at "a".

For every x in I not equal to a, if we have:

$\large\color{brown}{ g(x) \le f(x) \le h(x) \, \text{ } \\ \text{ and } \\ } \text{ } \\ \large\color{brown}{\eqalign{ \lim_{ x \to a} g(x) = \lim_{ x \to a} h(x) = L. \, } \text{ } \\ \\\text{ } \\ then \text{ } \\ \text{ } \\ \eqalign{ \lim_{x \to a} f(x) = L. } }$ When the limits on the upper bound and lower bound are the same, then the function in the middle is squeezed into having the same limit.

2. Examples

Example 1

$\large\color{green}{ \eqalign{ \lim_{ x \to 0} \frac{sin (x)}{x} = 1. } }$



We'll use the Squeeze theorem by establishing upper and lower bounds on sin(x)/x in an interval around 0;

Note the two functions cos(x) and sin(x)/x are even.
Thus, we'll just establish le bounds on the interval (0, π/2).

Let's compare the areas:

area of triangle POA ≤ area of sector POA ≤ area of triangle TOA.

That is:

sin(x)/2 ≤ x/2 ≤ tan(x)/2

2/sin(x) is positive. Hence:

1 ≤ x/sin(x) ≤ 1/cos(x)

so
$$\large\color{green}{ cos(x) \le \frac{sin(x)}{x} \le 1 }$$

Now let's take the limit:

cos(x) = cos(0) = 1, by continuity of cosine.

Hence

$\text{ } \\ \large\color{brown}{\eqalign{ \lim_{ x \to 0} cos(x) \le \lim_{ x \to 0} \frac{sin(x)}{x} \le \lim_{ x \to 0} 1 }} \text{ } \\ \text{ } \\ \text{ } \\ \large\color{brown}{\eqalign{ 1 \le \lim_{ x \to 0} \frac{sin(x)}{x} \le 1 }} \text{ } \\ \text{ } \\ \text{ } \\ \Large\color{blue}{ \eqalign{ \lim_{ x \to 0} \frac{sin (x)}{x} = 1. } }$

Example 2

$\text{ } \\ \large\color{green}{ \eqalign{ \lim_{ x \to 0} x cos(\frac{1}{x}) = 0. } } \text{ } \\ \text{ } \\ \text{ We have:} \\ \text{ } \\ \eqalign{ \lim_{ x \to 0} x = 0. } \text{ } \\ \text{ Because} \eqalign{ \lim_{ x \to 0} cos(\frac{1}{x})} \text{ does not exist, } \text{ we cannot use the multiplication rule. } \\ \text{ We use the squeeze theorem instead. } \\ \text{ } \\ \text{ Because cosine is always between - 1 and + 1, we have:} \\ \text{ } \\ -|x| \le x cos(\frac{1}{x}) \le |x| \text{ } \\ \text{ } \\ \text{ Then } \\ \text{ } \\ \large\color{brown} { \eqalign{ \lim_{ x \to 0} -|x| \le \lim_{ \; x \to 0} x cos(\frac{1}{x}) \le \lim_{ x \to 0} |x|\, } } \text{. so} \\ \text{ } \\ \text{ } \\ \Large\color{indigo}{ \eqalign{ \lim_{ x \to 0} \; x \; cos(\frac{1}{x}) = 0. } }$

Example 3

$\text{ } \\ \large\color{green}{ \eqalign{ \lim_{ x \to 0} (\frac{sin^2 x}{x}) = 0. } } \text{ } \\ \text{ } \\ \text{ We have:} \\ \text{ } \\ \eqalign{ \lim_{ x \to 0} (\frac{sin^2 x}{x}) = \lim_{ x \to 0} (\frac{sin x}{x}) \lim_{ x \to 0} (sin x) = 1 . 0 = 0 } \text{ } \\ \text{ } \\ \text{ } \\ \large\color{brown}{ \eqalign{ \lim_{ x \to 0} (\frac{sin^2 x}{x}) = 0. } }$

Example 4

$\text{ } \\ \large\color{green}{ \eqalign{ \lim_{ x \to 0} (\frac{1 - cos x}{x^2}) = \frac{1}{2} } } \text{ } \\ \text{ } \\ \text{We need to multiply by the conjugate of the numerator: } \\ \text{ } \\ \eqalign{ \lim_{ x \to 0} (\frac{1 - cos x}{x^2}) = \lim_{ x \to 0} ((\frac{1 - cos x}{x^2}) (\frac{1 + cos x}{1 + cos x})) = } \text{ } \\ \eqalign{ \lim_{ x \to 0} (\frac{1 - cos^2 x}{x^2 (1 + cos x)}) = \lim_{ x \to 0} (\frac{sin^2 x}{x^2 (1 + cos x)}) = } \text{ } \\ \eqalign{ (\lim_{ x \to 0} \frac{sin x}{x})^2 \lim_{ x \to 0} (\frac{1}{1 + cos x}) = } \text{ } \\ \eqalign{ 1^2 . \frac{1}{2} = \frac{1}{2}. } \text{ } \\ \text{ } \\ \text{ } \\ \Large\color{blue}{ \eqalign{ \lim_{ x \to 0} (\frac{1 - cos x}{x^2}) = \frac{1}{2} . } }$