# How to prove the continuity of

The function is called continuous if it displayed no jumps for small variations of the argument between these points. Graphically, this function is represented by a solid line without gaps.

## Instruction how to prove the continuity of

Step 1:

Proof of continuity at the point by means of so-called ε-Δ-argument. ε-Δ definition is: let x_0 belongs to the set X, then the function f (x) is continuous at x_0, if for any e gt; 0 there exists a Δ gt; 0 such that | x - x_0 | lt; Δ be | f (x) - f (x_0) | lt; e. The function is continuous on the set X, if it is continuous at every point. Another definition of continuity: the function f (x) is continuous at x = x_0, if the increment of the function at this point is infinitely small. Those. f (x) = f (x_0) + α (x), where α (x) - infinitesimal when x, tends to x_0.

Step 2:

Example 1: prove continuity of the function f (x) = x ^ 2 at x_0. Proof By definition, ε-Δ there exists e gt; 0 that | x ^ 2 - x_0 ^ 2 | lt; e. Give inequality to the form with an unknown quantity | x - x_0 |: | x ^ 2 - x_0 ^ 2 | = | X ^ 2 - 2 * x * x_0 + x_0 ^ 2 + 2 * x_0 * x - 2 * x_0 ^ 2 | = | (X - x_0) ^ 2 + 2 * x_0 * (x - x_0) | lt; e.

Step 3:

Solve the quadratic equation (x - x_0) ^ 2 + 2 * x_0 * (x - x_0) - ε = 0. Find the discriminant D = √ (4 * x_0 ^ 2 + 4 * ε) = 2 * √ (| x_0 | ^ 2 + ε). Then the root is equal to | x - x_0 | = (-2 * X_0 + 2 * √ (| x_0 | ^ 2 + ε)) / 2 = √ (| x_0 | ^ 2 + ε). Thus, the function f (x) = x ^ 2 is continuous for | x - x_0 | = √ (| x_0 | ^ 2 + ε) = Δ.

Step 4:

Some elementary functions are continuous throughout the domain (set of values of X): f (x) = C (constant); All trigonometric functions - sin x, cos x, tg x, ctg x and so forth.

Step 5:

Example 2: prove continuity of the function f (x) = sin x. Proof By the definition of continuity in its infinitesimal increment function record: Δf = sin (x + Δx) - sin x.

Step 6:

Convert the formula for trigonometric functions: Δf = 2 * cos ((x + Δx) / 2) * sin (Δx / 2). cos function is limited for x ≤ 0, and the limit of sin (Δx / 2) tends to zero, hence it is infinitesimal when Δx → 0. The product of a bounded function and infinitely maloq values, and hence the increase of the initial function Δf is also infinitesimal. Therefore, the function f (x) = sin x is continuous for any value of x.