Finding the range of values of $\frac {\sqrt{1+x^2}-1}{x}$ How do I find the range of the above expression, given that $x \ \in \mathbb{R} -\{0\}$
A seemingly useful method is substituting $x=\tan \alpha$
If $y$ be the given expression, then following my substitution 
$$y=\frac {1- | \cos \alpha |}{\sin \alpha}$$
Which can be simply dealt with by taking different cases for positive and negative values of $\cos \alpha $
Is there a simpler algebraic or calculus based method?
 A: Let $$y = \frac{\sqrt{1+x^2}-1}{x}$$
Then $xy + 1 = \sqrt{1+x^2} \Rightarrow x^2y^2 + 2xy + 1 = 1+x^2 \Rightarrow x^2(y^2-1) + 2xy = 0$.
So either $x=0$ (wich cannot be) or:
$$x = \frac{-2y}{y^2-1}$$
So $y$ can have any value except $\pm 1$,.
Note that $y$ can't be $0$ either because that would force $x=0$.
Since $y$ is a continuous function except when  $x = 0$ it means that the range of $y$ is either $(-\infty,-1)\cup (1,\infty)$ or $(-1,0)\cup (0,1)$.
A simple check tells you it's $(-1,0)\cup (0,1)$
A: Yes, there is. Why not differentiate it?
Letting $f(x)$ be your function, you'll have 
$$f'(x)=\frac{\sqrt{1+x^2}-1}{x^2\sqrt{1+x^2}}\gt 0.$$
So, we know $f(x)$ is strictly increasing. 
Also, you'll have 
$$\lim_{x\to\infty}f(x)=\lim_{x\to \infty}\left(\sqrt{\frac{1}{x^2}+1}-\frac 1x\right)=1,$$$$\lim_{x\to -\infty}f(x)=\lim_{x\to -\infty}\left(\color{red}{-1}\times\sqrt{\frac{1}{x^2}+1}-\frac{1}{x}\right)=-1,$$$$\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{(\sqrt{1+x^2}-1)'}{x'}=\lim_{x\to 0}\frac{x}{\sqrt{1+x^2}}=0.$$
Hence, the answer is $-1\lt y\lt 0\ \text{or}\ 0\lt y\lt 1.$
A: From the plot of the function (an S-shaped curve with two horizontal asymptotes), it is clear that the range is $(-1,1)$.
This can be proven from inequalities.
If
$$x\ge0,$$
$$(x+1)^2\ge x^2+1\ge1,$$
$$x\ge\sqrt{x^2+1}-1\ge0,$$
$$1\ge\frac{\sqrt{x^2+1}-1}x\ge0.$$
These bounds are reached for $x=\infty$ and $x=0$.
As the function is odd, by symmetry the full range is indeed $(-1,1)$.
