Squeeze Theorem for Function Limits

Let $ h(x) $, $ f(x) $, and $ g(x) $ be functions defined on the same domain, possibly with the point $ x_0 $ excluded. Suppose that for every $ x \neq x_0 $, $$ h(x) \le f(x) \le g(x) $$ and that $$ \lim_{x \to x_0} h(x) = \lim_{x \to x_0} g(x) = \ell $$ Then, $$ \lim_{x \to x_0} f(x) = \ell $$

The squeeze theorem is a fundamental tool for evaluating limits. It works by comparing a function with two others that bound it from below and above.

If $ f(x) $ is confined between two functions that converge to the same limit, then $ f(x) $ must converge to that limit as well.

Intuitively, $ f(x) $ is "trapped" between the bounding functions and therefore cannot approach any value different from their common limit.

Note. The geometric interpretation is immediate. As $ x $ approaches $ x_0 $, both $ h(x) $ and $ g(x) $ become arbitrarily close to $ \ell $. Because $ f(x) $ always lies between them, it too must approach $ \ell $.

When is it used?

The squeeze theorem is particularly effective when the limit of $ f(x) $ is difficult to compute directly, yet appropriate comparison functions with known limits are readily available.

A Practical Example
Proof

A Practical Example

Consider a function $ f(x) $ whose values, in a neighborhood of $ 0 $, satisfy

$$ -x^2 \le f(x) \le x^2 $$

The bounding functions share the same limit:

$$ \lim_{x \to 0} (-x^2) = 0, \qquad \lim_{x \to 0} x^2 = 0 $$

Hence, by the squeeze theorem,

$$ \lim_{x \to 0} f(x) = 0 $$

This result follows without requiring the explicit analytic form of $ f(x) $.

Note. The squeeze theorem applies equally to limits as $ x \to +\infty $ or $ x \to -\infty $. The limits of the comparison functions are not required to be finite.

Example 2

Consider the function

$ f(x) = x^2 \sin \left(\frac{1}{x}\right) $, for $ x \neq 0 $.

We evaluate the limit as $ x \to 0 $:

$$ \lim_{x \to 0} f(x) $$

From the elementary property of the sine function

$$ -1 \le \sin \left(\frac{1}{x}\right) \le 1 $$

Multiplying each term by $ x^2 $,

$$ -x^2 \le x^2 \sin \left(\frac{1}{x}\right) \le x^2 $$

Since $ x^2 \ge 0 $, the inequalities preserve their direction.

Now take limits:

$$ \lim_{x \to 0} (-x^2) \le \lim_{x \to 0} x^2 \sin \left(\frac{1}{x}\right) \le \lim_{x \to 0} x^2 $$

The limits of the comparison functions are

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

$$ \lim_{x \to 0} x^2 = 0 $$

Substituting these values,

$$ 0 \le \lim_{x \to 0} x^2 \sin \left(\frac{1}{x}\right) \le 0 $$

Therefore,

$$ \lim_{x \to 0} x^2 \sin \left(\frac{1}{x}\right) = 0 $$

Although $ \sin(1/x) $ oscillates indefinitely between -1 and 1, the factor $ x^2 $ dominates the behavior and drives the product to zero.

Proof

Let $ \varepsilon > 0 $ be arbitrary.

Assume that

$$ h(x) \le f(x) \le g(x), \qquad x \neq x_0 $$

and that

$$ \lim_{x \to x_0} h(x) = \ell, \qquad \lim_{x \to x_0} g(x) = \ell $$

By the definition of limit, there exists a neighborhood $ I_1 $ of $ x_0 $ such that

$$ |h(x) - \ell| < \varepsilon \quad \text{for all } x \in I_1, \; x \neq x_0 $$

Equivalently,

$$ \ell - \varepsilon < h(x) < \ell + \varepsilon $$

Similarly, there exists a neighborhood $ I_2 $ of $ x_0 $ such that

$$ |g(x) - \ell| < \varepsilon $$

that is,

$$ \ell - \varepsilon < g(x) < \ell + \varepsilon $$

Let $ I = I_1 \cap I_2 $, which is itself a neighborhood of $ x_0 $.

For every $ x \in I $, with $ x \neq x_0 $

$$ \ell - \varepsilon < h(x) \le f(x) \le g(x) < \ell + \varepsilon $$

Hence,

$$ \ell - \varepsilon < f(x) < \ell + \varepsilon $$

which is equivalent to

$$ |f(x) - \ell| < \varepsilon $$

Since this holds for every $ \varepsilon > 0 $, we conclude

$$ \lim_{x \to x_0} f(x) = \ell $$

As required.

And so on.