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.

$\require{cancel}$
 Multiply numerator and denominator by the conjugate of the numerator: $$\sqrt{x+1} + 1$$ then evaluate the limit. 
When we multiply by the conjugate, recall how we factor the difference of squares: $$(\sqrt a - b) \cdot (\sqrt a + b) = (\sqrt a)^2 - b^2 = a - b^2$$
$$\dfrac {\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} = \dfrac {(x + 1) - 1 }{x(\sqrt{x+1} + 1)}= \dfrac {\cancel{x}}{\cancel{x}(\sqrt{x+1} + 1)} = \dfrac 1{\sqrt{x + 1} + 1}$$
Now we need only to evaluate $$\lim_{x \to 0} \dfrac 1{\sqrt{x + 1} + 1}$$
I trust you can do that.
