This problem is from the book
https://www.amazon.com/Vector-Calculus-Dover-Books-Mathematics/dp/0486466205
It's on page 24.
Prove that when $a \gt 0$ and $|h| \lt a$, then $\sqrt{a^2+h}\ $ differs from $a + (1/2)(h/a)$ by an amount less than $h^2 / (8a^3)$.
I feel there's something wrong here (several things actually).
I think the statement should say $|h| \lt a^2$.
Why? Well, if e.g. $a = 1/2$ and $h = -1/3$, then $\sqrt{a^2+h}\ $ is not even defined.
Or maybe the problem statement should just say $a > 1$. Then we don't have such a problem because $|h| < a$ implies $|h| < a^2$.
Otherwise, I think the solution is "kind of" clear to me but not quite. There's another problem here as well (with the solution).
I define $f(x) = \sqrt{x}$
Then I let $b = a^2$. I then use the Taylor formula and I write
$f(b+h) = f(b) + (h/1!)f'(b) + (h/2!)f''(b + \theta h)$
where $\theta \in (0,1)$
Here we need to have that $|h| < b\ $, hence my worry from above.
OK, let's say we have that for $h$ i.e. $|h| < b\ $.
So I obtain that
$$\sqrt{a^2+h} = a + (1/2a)h + (-1/8)h^2\frac{1}{(a^2+\theta h)^{3/2}}$$
If $h > 0$ everything is fine.
I mean, in that case I can obtain what I need for the last term, namely that
$\frac{1}{(a^2+\theta h)^{3/2}} < 1/a^3$
But if $h < 0$, can we claim than $a^2+\theta h > a^2$ ?
We can't, right?! If we can't then we can't get the estimate we need for the last term, right?
Say e.g. we take $a=0.5, b=0.25, h=-0.20$
Is the statement actually true in that case?
I wrote a small program and in that case it doesn't seem to be true.
Of course in the numeric example given, there is no such problem. I take a = 10, h=8, and also we have $a > 1$. So none of the problems mentioned above exist here.
But as the problem is stated, I don't think what it claims is true.
I wonder if my thoughts are correct and if so, how can the problem statement be fixed?! To me the most obvious way to fix the problem statement is probably to say that $a > 1$ and that $ 0 < h < a$. But the way it's stated, it's too general. I think it's quite flawed.