Find the range of possible values of $\sqrt{a^2+a+1}-\sqrt{a^2-a+1}$ 
Let $x=\sqrt{a^2+a+1}-\sqrt{a^2-a+1},x\in \mathbb R$. Find range of possible values of $x$.

I tried drawing the graph and obtained this:
Through which the answer came out to be $(-1,1)$.
What should be the procedure through algebra?
 A: Since $f(-a)=-f(a)$, we only need to find the range of $f(a)$ for $a\ge0$. Now, we have $$g(a):=f(a)^2=2a^2+2-2\sqrt{(a^2+1)^2-a^2}=2a^2+2-2\sqrt{a^4+a^2+1},$$
so
$$g'(a)=4a-2\frac{4a^3+2a}{2\sqrt{a^4+a^2+1}}=4a\left(1-\frac{a^2+\frac12}{\sqrt{(a^2+\frac12)^2+\frac34}}\right)\ge0,$$ and hence $g(a)$ (and hence also $f(a)$) increases on $[0,\infty)$. Note also that $g'(a)>0$ whenever $a>0$ so $f(a)$ is strictly increasing.
Since $f$ is continuous, the range of $f$ on $[0,\infty)$ is $[0,A)$ where $$A=\lim_{a\to\infty}f(a)=\lim_{a\to\infty}\frac{2a}{\sqrt{a^2+a+1}+\sqrt{a^2-a+1}}=1.$$

P.S. an alternative method, which is probably way better
Let $h(a)=a+1-\sqrt{a^2+a+1}$ so that $g(a)=2h(a^2)$. Now, we wish to prove $h(a)<\frac12$ for $a>0$, or equivalently, $\sqrt{a^2+a+1}>a+\frac12$. This follows from $a^2+a+1=\sqrt{(a+\frac12)^2+\frac34}$.
Now, again by computing $\lim_{a\to\infty}f(a)$ you arrive at the same conclusion.
A: Multiply your expression by
$$\frac{  \sqrt{a^2+a+1}+\sqrt{a^2-a+1} } {\sqrt{a^2+a+1}+\sqrt{a^2-a+1}}$$
To get
$$\frac{2a}{ \sqrt{a^2+a+1}+\sqrt{a^2-a+1}}.$$
For $a$ positive, divide top and bottom by $a$ and push
the $1/a$ inside the square roots:
$$\frac{2}{ \sqrt{ 1+ \frac{1}{a} +\frac{1}{a^2} } + 
\sqrt{ 1 - \frac{1}{a} +\frac{1}{a^2} }}.$$
Now you can see that as $a$ gets very large, the above
fraction gets close to
$$\frac{2}{1+1} = 1.$$
For $a$ negative, do the same thing, but when you push the $1/a$ inside the square roots, you have to leave the negative outside.  The expression tends to $-1$.
Also, looking at the last expression, you can work out that the denominator is larger than the absolute value of the numerator, so the range is $(-1,1)$.
A: Square both sides to get
$$
x^2 = 2(a^2 + 1  - \sqrt{a^4 + a^2 + 1})
$$
We can then use $a^4 + a^2 + 1 > (a^2 + 1/2)^2$ to see that $x^2 < 1$. Now rearrange and square again to get
$$
(x^2 - 2a^2 - 2)^2 = 4(a^4 + a^2 + 1).
$$
Assuming $|x| < 1$, we can solve for $a$ to get
$$
a =x\sqrt{\frac{1- (x/2)^2}{1-x^2}}.
$$
This is defined for every $x \in (-1,1)$, so $x(a)$ covers the entire range $(-1,1)$.
