Why is $\frac{\sqrt{x+1}-1}{x}$ equal to $\frac{1}{\sqrt{x+1}+1}$?

I'm working with the expression $$\frac{\sqrt{x+1} - 1}{x}.$$ According to Wolfram Alpha "alternate form" section (http://www.wolframalpha.com/input/?i=%28%28x%2B1%29%5E1%2F2-1%29%2Fx) it is equal to $$\frac{1}{\sqrt{x+1} + 1}.$$ How can I go from one expression to the other?

$$\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}-1}{x}\times\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\frac{x}{x\left(\sqrt{x+1}+1\right)}=\frac{1}{\sqrt{x+1}+1}$$

• (+1), notice however that while the domain of the first is $\mathbb{R}\backslash\{0\}$, the domain of the second is $\mathbb{R}$ (I guess it could be important depending on what you do :?) – servabat Jan 22 '15 at 19:19
• I'd check the domain restrictions, @servabat. Also, note that the second equality follows from the difference of two squares. – The Chaz 2.0 Jan 22 '15 at 19:20
• @servabat - For the function to be real-valued (which I assume is supposed to be the case, although it's not stated explicitly), its domain must be $\{x\mid x\ge-1\}$, not $\mathbb{R}$. Using L'Hopital's Rule, one finds $f(x)=0$ as $|x|\to0$, i.e., $\{0\}$ is included in the function's domain. – Mico Jan 22 '15 at 21:42
• Was not specified, and I think ones should always consider the largest (so here $\mathbb{C}$) when not specified. However, because the function has a limit in a point doesn't mean that that function is defined in that point and yet, while you don't specify it, it is not defined. So because you find a limit in a point doesn't basically mean it is in the domain. By the way, there was no need of L'Hospital or anything else because basically you have an equality so the limit is the value of the function at the point, which you can easily evaluate in the first expression. – servabat Jan 22 '15 at 21:54
• Because in the second expression the domain is narrowed doesn't mean it will change the limit. – servabat Jan 22 '15 at 21:55

Multiply the top and bottom by the conjugate $(x+1)^{1/2}+1$

Hint $\$ Rationalize the $\, \color{#c00}{numerator},\,$ in exactly the same way one rationalizes denonominators, i.e. multiply both by the "conjugate" $\ \overline{1-\sqrt{x+1}} \,=\, 1+\sqrt{x+1}\,$ of the numerator.

Remark $\$ Though not used as frequently as rationalizing the denominator, this does prove useful in various contexts, e.g. specializing the quadratic formula when the leading coeff $\,a\to 0.$

$$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\ =\ \dfrac{2c}{-b \pm \sqrt{b^2-4\:a\:c}}$$

As $\,a\to 0,\,$ the latter gives the root $\,x = -b/c\,$ of $\,bx+c\,\ (= ax^2\!+bx+c\,$ when $\,a=0).$