$f(x) = \frac{\sqrt{1 + x} - 1}{x}$ for $x \ne 0$, and $f(0) = k$. For which $k$ is $f$ continuous at $x = 0$?

Explanation

1. 01

   Multiply above and below by $\sqrt{1 + x} + 1$: the expression becomes $\frac{1}{\sqrt{1 + x} + 1}$.

2. 02

   At $x = 0$ that is $\frac{1}{2}$.