# How do you prove that h(x) = sqrt(2x - 3) is continuous as x=2?

Apr 12, 2018

A function $f \left(x\right)$ is said to be continuous at $a$ if ${\lim}_{x \rightarrow a} f \left(x\right) = f \left(a\right)$

So in this case, our $f \left(x\right)$ is $\sqrt{2 x - 3}$ and $a$ is $2$. Compute the limit, then verify that it equals $h \left(2\right)$:

$\textcolor{w h i t e}{=} {\lim}_{x \rightarrow 2} \sqrt{2 x - 3}$

$= \sqrt{2 \left(2\right) - 3}$

$= \sqrt{4 - 3}$

$= \sqrt{1}$

$= 1$

Here's $h \left(2\right)$:

$h \left(2\right) = \sqrt{2 \left(2\right) - 3}$

$\textcolor{w h i t e}{h \left(2\right)} = \sqrt{4 - 3}$

$\textcolor{w h i t e}{h \left(2\right)} = \sqrt{1}$

$\textcolor{w h i t e}{h \left(2\right)} = 1$

Since the limit and the function are equal, the function $h \left(x\right)$ is continuous at $x = 2$. Hope this helped!