Merging two functions So I have two fuctions like this :-
$f(x) = (x/5)^2$
and $g(x) = \sqrt{(x/5)}$
and a third fuction as a combination of both
$ h(x) =  \Biggl[ { }^{ x\; \lt \; 5 : \; f(x) }_{ x \;\ge \; 5: \; g(x)}\Biggr] $
When I put $x =5$ in the first function I get $f(5) = 1$ and in the second one I get $g(5) = 1\;$. That means h(x) is countinous at $x =5$ is there any way I can converge $h(x)$ into a single function??
 A: 
Is there any way I can [combine] $h(x)$ into a single function?

You just did. What you just wrote is a completely acceptable definition of one single function $h$. 
However, you may be wondering something. Namely, your function is written in a piecewise form (it's defined separately for $x < 5$ and $x > 5$), and you may be wondering if it's possible to write your function in a non-piecewise form.
The answer is no, not really. The function $h$ is sort of an "inherently piecewise function"; in order to define $h$, you pretty much need to either define it as a piecewise function, or define it in terms of another piecewise function (such as the absolute value function). 
A: You mean one "formula", not one "function". drhab showed one way to do that, using the functions $1_{x< 5}$, the function that is equal to one if x< 5, 0 otherwise and $1_{x\ge 5}$, the function that is 1 if $x\ge 5$, 0 otherwise.
A different way of writing this is to use the "Heaviside step function", H(x), which is defined to be 0 for x< 0, 1 for $x\ge 0$. H(x-a) is 0 for x< a, 1 for $x\ge a$.   A function that has value f(x) for x< a and g(x) for $x\ge a$ is f(x)+ (g(x)- f(x))H(x- a).  When x< a, x- a< 0, H(x- a)= 0, so we have f(x)+ (g(x)- f(x))(0)= f(x).  When $x\ge a$, H(x- a)= 1, so we have f(x)+ (g(x)- f(x))(1)= g(x).
Here, you want $(x/5)^2+ (\sqrt{x/5}- (x/5)^2)H(x- 5)$.
A: Others have shown how you may be able to disguise the piecewise nature of your function e.g. by using the Heaviside step function.  If that is too exotic for you then similar tricks may be possible with more familiar functions e.g. absolute value.  
An indication that a very nice answer is unlikely is that although your function is continuous at $5$, it is not differentiable there.  The derivative (slope) abruptly changes at $5$.  Pretty much any nice answer will be differentiable as well as continuous.  
To give a more exact answer, we would need to know what you consider acceptable.  
A: $$h(x) = \left ( \frac 1 {10} \left (x - 5 - \sqrt {\strut (x - 5)^2} \right ) + 1 \right )^2 \sqrt {\frac 1 {10} \left (x - 5 + \sqrt {\strut (x - 5)^2} \right ) + 1}$$
