How to evaluate this limit: $\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac12$?

Question

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.

Answer 1

$\require{cancel}$ 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.

Answer 2

Method 1 (basic)
$$\frac{\sqrt{x+1} - 1}{x} \stackrel{\sqrt{x+1}^2 = |x+1|}{=} \frac{|x + 1| - 1}{x (\sqrt{x+1} + 1)} \stackrel{x+1 \geq 0, \text{ for well-def.}}{=} \frac{1}{\sqrt{x + 1} + 1} \to \frac{1}{1+1} = \frac{1}{2}$$ as $x\to 0$, because $x\mapsto \sqrt{x}$ is continuous (and so the limit can be "used as input")
Method 2 (derivative)
$$\frac{\sqrt{x+1} - 1}{x} = \frac{f(1 + x) - f(1)}{x}$$ where $f(y) = \sqrt{y}$, thus the limit is $$f'(1) = \frac{1}{2 \sqrt{1}} = \frac{1}{2}$$ Method 3 (l'Hospital)
Since the form is $\frac{0}{0}$, l'Hospital can be applied and gives $$\lim_{x\to 0} \frac{\sqrt{x+1} - 1}{x} = \lim_{x\to 0} \frac{\frac{1}{2\sqrt{x+1}}}{1} = \lim_{x\to 0} \frac{1}{2\sqrt{x + 1}} = \frac{1}{2 \sqrt{1}} = \frac{1}{2}$$