Evaluate a limit by rationalizing the numerator

Multiply the numerator and denominator by the conjugate $\sqrt{x+1}+1$: $\frac{\sqrt{x+1}-1}{x} \cdot \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1} = \frac{(x+1) - 1}{x(\sqrt{x+1}+1)} = \frac{x}{x(\sqrt{x+1}+1)}.$ This uses the identity $(a-b)(a+b) = a^2 - b^2$ to eliminate the square root in the numerator.

Cancel the common factor of $x$ (valid since $x \neq 0$ near the limit) to get $\frac{1}{\sqrt{x+1}+1}.$ This function is continuous at $x=0$, so direct substitution now gives $\frac{1}{\sqrt{0+1}+1} = \frac{1}{2}$.