How to evaluate this limit: $\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac12$? How do I evaluate the limit of $$\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $\frac{1}{2}$, but I got this question wrong.  I think you have to multiply by the conjugate of the numerator?  Can someone explain the steps to this, I must be doing something wrong when I multiply.

Alternative solution: Write $ t= \sqrt{x+1}$ then $x=t^2-1$ and we get $$\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \lim_{t\to 1}\frac{t-1}{t^2-1} =\lim_{t\to 1}\frac{t-1}{(t-1)(t+1)}=\frac12$$
