Theorem on the Preservation of Sign for the Limit of a Function

Let f(x) be a function defined and continuous in a neighborhood of the point x₀. If f(x₀) > 0, then there exists a number δ > 0 such that f(x) > 0 for all x in the interval (x₀ - δ, x₀ + δ).

Example

Consider the function:

$$ f(x) = \frac{1}{x} $$

At the point x₀ = 2, the function evaluates to f(x₀) = 0.5:

$$ f(2) = 0.5 $$

Let’s look at a neighborhood around x₀ = 2 with δ = 0.2:

$$ (x_0 - \delta, \, x_0 + \delta) $$

$$ (2 - 0.2, \, 2 + 0.2) $$

$$ (1.8, \, 2.2) $$

Within the interval (1.8, 2.2), the function f(x) remains strictly positive:

$$ f(1.8) = \frac{1}{1.8} = 0.55 \\ f(2.2) = \frac{1}{2.2} = 0.45 $$

Graphically, this can be illustrated as follows:

Note. For any x₀ such that f(x₀) > 0, it is always possible to find some δ > 0 that defines a neighborhood in which f(x) > 0 for all x ∈ (x₀ - δ, x₀ + δ). 

Proof

Since the function is positive at the point x₀:

$$ f(x_0) > 0 $$

We choose:

$$ \epsilon = \frac{f(x_0)}{2} $$

By the definition of the limit, there exists a δ > 0 such that:

$$ |f(x) - f(x_0)| < \epsilon $$

which is equivalent to:

$$ |f(x) - f(x_0)| < \frac{f(x_0)}{2} $$

This leads to the double inequality:

$$ -\frac{f(x_0)}{2} < f(x) - f(x_0) < \frac{f(x_0)}{2} $$

Adding f(x₀) to all sides gives:

$$ f(x_0) - \frac{f(x_0)}{2} < f(x) < f(x_0) + \frac{f(x_0)}{2} $$

Since f(x₀) - f(x₀)/2 simplifies to f(x₀)/2:

$$ f(x_0) - \frac{f(x_0)}{2} = \frac{f(x_0)}{2} $$

And because f(x₀) > 0, it follows that f(x₀)/2 > 0. Therefore, f(x) remains strictly positive in the neighborhood:

$$ 0 < \frac{f(x_0)}{2} < f(x) < \frac{3f(x_0)}{2} $$

And so on.