Question

Let the function f(x) be continuous at a point X_o. Assume that f(X_o) > 2. Prove that there exists a neighborhood of X_o such that f(x) > 2 for every x in this neighborhood.

Answer 1

Conditions

Let the function $f(x)$ be continuous at a point $X_{-}o$. Assume that $f(X_{-}o) \& gt; 2$. Prove that there exists a neighborhood of $X_{-}o$ such that $f(x) \& gt; 2$ for every $x$ in this neighborhood.

Solution

$$\forall \varepsilon > 0 \ \exists \delta = \delta(\varepsilon) > 0 \ \forall x: |x - x_0| < \delta \ |f(x) - f(x_0)| < \varepsilon$$ $$f(x_0) = 2 + k, k > 0$$ $$|f(x) - f(x_0)| < \varepsilon$$ $$2 + k - \varepsilon < |f(x)| < 2 + k + \varepsilon$$

As we know, these claims are true for all $\varepsilon > 0$. And for those $\varepsilon$, which are:

$$0 < \varepsilon < k$$ $$\exists \delta = \delta(\varepsilon) > 0 \ \forall x: |x - x_0| < \delta \ |f(x) - f(x_0)| < \varepsilon$$

And

$$2 < 2 + k - \varepsilon < |f(x)| < 2 + k + \varepsilon$$

Q.E.D.