The conjugate trick at zero

Idea. The numerator $\sqrt{1+x}-1$ and the denominator $x$ both tend to $0$. The identity $(A-B)(A+B)=A^2-B^2$ removes the square root and creates a factor $x$ that we can cancel.

Computation. For $x>-1$, $x\neq0$,
$$\begin{aligned} \frac{\sqrt{1+x}-1}{x}&=\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x\,(\sqrt{1+x}+1)} \\ &=\frac{(1+x)-1}{x\,(\sqrt{1+x}+1)} \\ &=\frac1{\sqrt{1+x}+1}. \end{aligned}$$
As $x\to0$, $\sqrt{1+x}\to1$, so the limit is $\dfrac1{1+1}=\dfrac12$.

Remark. This limit is the derivative of $t\mapsto\sqrt t$ at $t=1$, which is $\frac1{2\sqrt1}=\frac12$. Recognising a derivative is often the fastest way to compute a limit.