The Squeeze Theorem, also called the Sandwich Theorem, is an important idea in calculus. It helps us find the limit of a function by comparing it to two simpler functions. This is especially helpful when the function is hard to evaluate directly. Let’s look at some examples that show how the Squeeze Theorem works.

1. Limit of (\frac{\sin(x)}{x}) as (x) Approaches 0

One classic example uses the function (\frac{\sin(x)}{x}). As (x) gets closer to 0, both (\sin(x)) and (x) also get closer to 0, making it hard to solve directly. But we can use the Squeeze Theorem here.

• When (0 < x < \frac{\pi}{2}): $$\cos(x) < \frac{\sin(x)}{x} < 1$$
• As (x) gets close to 0:
  • (\cos(x)) gets closer to (1)
  • The upper limit stays at (1)

So, by using the Squeeze Theorem:

$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$

2. Limit of (\frac{1 - \cos(x)}{x^2}) as (x) Approaches 0

Another example looks at the limit of (\frac{1 - \cos(x)}{x^2}) as (x) approaches 0.

• We know that (1 - \cos(x)) can be estimated by (\frac{x^2}{2}) when (x) is close to 0: $$0 \leq \frac{1 - \cos(x)}{x^2} < \frac{x^2/2}{x^2} = \frac{1}{2}$$

As (x) approaches 0, both limits get closer to 0, giving us:

$$\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = 0$$

3. Limit of (x \sin\left(\frac{1}{x}\right)) as (x) Approaches 0

Another interesting case is the limit of (x \sin\left(\frac{1}{x}\right)) as (x) approaches 0.

• The sine function moves between -1 and 1: $$-1 < \sin\left(\frac{1}{x}\right) < 1$$
• When we multiply everything by (x) (this works when (x) is close to 0 and positive): $$-x < x \sin\left(\frac{1}{x}\right) < x$$

As (x) approaches 0, both (-x) and (x) get closer to 0:

$$\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$$

Conclusion

The Squeeze Theorem is a great tool for finding limits in calculus, especially for functions that are hard to evaluate directly. It helps make sense of functions that wiggle around and is key to understanding limits. Examples like (\lim_{x \to 0} \frac{\sin(x)}{x}), (\lim_{x \to 0} \frac{1 - \cos(x)}{x^2}), and (\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)) show why this theorem is important in calculus. Math can help us understand these ideas better by comparing different functions!