# Proving uniformly continuity of 2 functions

Domain: $$(0, \infty)$$

I have $$2$$ functions:

$$f(x) = \sqrt{x}, \quad g(x) = x \cdot \sin(1/x)$$

The answers say that $$f(x)$$ is uniformly continuous because at $$0$$ it has a finite limit and in the $$\infty$$ its derivative is bounded.

For $$g(x)$$, it has finite limits at the boundaries of the interval, namely at $$0$$ and $$\infty$$ and therefore it is uniformly continuous.

Can someone explain to me how those facts prove that the functions are uniformly continuous?

I am familiar with the formal definition and with the fact that if the derivative is bounded in the interval than the function is uniformly continuous (for continuous functions).

• You should clearly specify the domain of these functions. – zhw. May 23 at 17:31

Consider a continuous function $$f:(a,b) \rightarrow \mathbb{R}$$ with $$-\infty \le a < b \le \infty$$. What the answers uses is that when $$\lim\limits_{x \rightarrow a} f(x)$$ and $$\lim\limits_{x \rightarrow b} f(x)$$ exists, then $$f$$ is uniform continuous. To understand this, note that by existence of $$f := \lim\limits_{x \rightarrow a} f(x)$$ we can find for all $$\varepsilon >0$$ some $$b' such that for all $$x\ge b'$$ we have $$|f(x) - f| < \varepsilon$$. Therefore $$|f(x) - f(y)| < 2\varepsilon$$ for all $$x,y\ge b'$$. Analogously we can find some $$a' > a$$. Now $$f$$ is uniform continuous on the compact set $$[a',b']$$ and you combine this with previous observation to get uniform continuouty on (a,b). Take a look here.
Now in you first example, as you said, $$|f'(x)|$$ is bounded on $$[1, \infty)$$ and therefore $$f$$ is uniform continuous on $$[1, \infty)$$. Moreover, by the above reasoning $$f(x)$$ is also uniform continuous on (0,1). Here, we can also use that $$[0,1]$$ is compact and hence $$f$$ is uniform contionuous.