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.

$x>-1$.
$x= (x +1)- 1= \sqrt{(x+1)}^2-1^2$
$=(\sqrt{(x+1)}-1)(\sqrt{x+1}+1)$.
$\dfrac{\sqrt{(x+1)}-1}{x} =\dfrac{x}{(\sqrt{x+1}+1)x}.$
Take the limit $x \rightarrow 0$.
