We can't expand this in a general case since it's very difficult to find power expansions for any case. However, we can estimate it in the limit of one being far larger than the other and then discuss that case.
We will assume that #x# is much greater than #h#. Therefore,
#sqrt(x+h) approx sqrtx * sqrt(1 + (h/x)) #
Now, we know that #h/x# is a very small value, so we can expand this using Taylor expansion. We need to find the #n#th derivative at #u=0# of #f(u) = sqrt(1+u)#. This is simple to find
#f(u) = (1+u)^(1/2) implies f(0) = 1 #
#f'(u) = 1/2 * (1+u)^(-1/2) implies f'(0) = 1/2 * 1#
#f''(u) = -1/2 * 1/2 * (1+u)^(-3/2) implies f''(0) = -1/2 * 1/2 * 1 = -1/4#
Imagining continuing this, it is clear that we have
#f^(n) (u) = 1/2 * (-1)/2 * ... (3-2n)/2 (1+u)^(1/2-n)#
#= -(-1/2)^n * (1 * 3 * 5 * ... (2n-3)) * (1+u)^(1/2-n)#
This middle value takes on the special character of a double factorial, hence for #n > 2#,
#f^(n) (0) = (-1)^(n+1)2^-n (2n-3)!!#
This gives us the expansion via Taylor's formula:
#sqrt(1 + u) = 1 + 1/2 u- 1/8 u^2 - sum_(n=3)^(infty) (-1/2)^n ((2n-3)!!)/(n!) u^n #
From this, we can recover the answer to the original from this:
#sqrt(x+h) approx sqrt(x) * sqrt(1 + (h/x)) = sqrt(x) * f(h/x) #
#approx sqrt(x) * [ 1 + 1/2 (h/x)- 1/8 (h/x)^2 - sum_(n=3)^(infty) (-1/2)^n ((2n-3)!!)/(n!) (h/x)^n ] #
Using just the first four terms, we get
#sqrt(x+h) approx x^(1/2) + 1/2 hx^(-1/2) - 1/8 h^2 x^(-3/2) + 1/16 h^3x^(-5/2) #
To see how accurate this is, let's say #x = 225# and #h = 64#. Therefore #sqrt(x+h) = 17#. Plugging in these numbers to the above formula,
#sqrt(x+h) approx 15 + 1/2 * 64 / 15 - 1/8 * 64^2 / (15^3) + 1/16 * 64^3 / (15^5) = 17.003 #
which is quite good already.
