# Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval.

$f(x) = x + \sqrt{x-4}, [4, \infty]$

This question aims to explain the concepts of continuity in functions, the difference between continuous and discontinuous functions, and understand the properties of limits.

When a continuous variation of the argument asserts a constant variation in the value of the function, It is called a continuous function. Continuous functions have no sharp changes in value. In continuous functions, a small change in the argument produces a small change in its value. Discontinuous is a function that is not continuous.

When a function approaches a number it is called the limit. For example a function $f(x) = 4(x)$, and the limit of the function f(x) is $x$ approaches $3$ is $12$, symbolically, it is written as;

$\underset{x \rightarrow 3}{lim} f(x) = 4(3) = 12$

Given that the function $f(x) = x + \sqrt{x-4}$ is defined on the interval $[4, \infty]$.

For $a > 4$ we have:

$\underset{x \rightarrow a}{lim} \space f(x) = \underset{x \rightarrow a}{lim} \space (x+ \sqrt{x-4})$

$=\underset{x \rightarrow a}{lim} \space x+\underset{x \rightarrow a}{lim} \space (\sqrt{x-4})$

$= \underset{x \rightarrow a}{lim} \space x+ \sqrt{\underset{x \rightarrow a}{lim} \space (x-4)}$

$=\underset{x \rightarrow a}{lim} \space x+ \sqrt{\underset{x \rightarrow a}{lim} \space x-\underset{x \rightarrow a}{lim} \space 4}$

$= a + \sqrt{a-4}$

$f(a)$

So the $\underset{x \rightarrow a}{lim} \space f(x) = f(a)$ for the all values of $a>4$. Therefore $f$ is continuous at $x=a$ for every $a$ in $(4, \infty)$.

Now checking at $\underset{x \rightarrow 4^+}{lim} \space f(x)$:

$\underset{x \rightarrow 4^+}{lim} \space f(x) = \underset{x \rightarrow 4^+}{lim} \space (x + \sqrt{x – 4})$

$= 4+\sqrt{4-4}$

$= 4+0$

$= 4$

$= f(4)$