Hello, h4hv4hd4si4n!
Remove the discontinuity: .$\displaystyle f(x) \;=\;\frac{x4}{\sqrt{x}  2} $
Since $\displaystyle f(4) \:=\:\frac{44}{\sqrt{4}2} \:=\:\frac{0}{0}$, there is a discontinuity at $\displaystyle x = 4.$ Code:

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 4
It is the upper half of a "horizontal" parabola
. . with a hole at (4, 4).
Rationalize: .$\displaystyle \frac{x4}{\sqrt{x}2}\cdot\frac{\sqrt{x} + 2}{\sqrt{x}+2} \;=\;\frac{(x4)(\sqrt{x} + 2)}{x  4} \;=\;\sqrt{x} + 2$
Hence: .$\displaystyle \lim_{x\to4}f(x) \;=\;\lim_{x\to4}(\sqrt{x}+2) \;=\;4$
Therefore: .$\displaystyle f(x) \;=\;\bigg\{\begin{array}{ccc}\frac{x4}{\sqrt{x}2} & & x \neq 4 \\ 4 & & x = 4 \end{array}$
. . and this "fills in the hole" . . . The function is continuous.