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?

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.
