Prove, by using the definition of continuity, that
$f(x) = \dfrac{5}{3}$$\sqrt x $ $+$ $3$
is continuous at the point $x=7$. Definition I have to use is as follows:
For any $\epsilon$>0, there exists $\delta>0$ such that: whenever $|x-x_0|<\delta$, we have $|f(x)-f(x_0)|<\epsilon$
Can someone please help me out here?
The thinking that follows applies to the general case where we prove that $x \mapsto \sqrt{x}$ is continuous everywhere in its domain.
We have $|\frac{5}{3}\sqrt{x} + 3 - \frac{5}{3}\sqrt{7} - 3| = \frac{5}{3}|\sqrt{x} - \sqrt{7}| = \frac{5}{3}\frac{|x-7|}{\sqrt{x}+\sqrt{7}}$ for all $x \geq 0$. If in addition we have $|x-7| < 1$, then $6 < x < 8$, so $\sqrt{6} < \sqrt{x} < \sqrt{8}$, and hence $\frac{|x-7|}{\sqrt{x}+\sqrt{7}} < \frac{|x-7|}{\sqrt{6}+\sqrt{7}}$. Given any $\varepsilon > 0$, we have $\frac{|x-7|}{\sqrt{6}+\sqrt{7}} < \varepsilon$ if in addition we have $|x-7| < \varepsilon (\sqrt{6}+\sqrt{7})$. So taking $\delta := \min \{ 1, \varepsilon (\sqrt{6}+\sqrt{7}) \}$ suffices.