How to find the left and right side limits of $f(x)=\frac{\sqrt{|1+x|}-1}{x}, f:\mathbb R \to \mathbb R$

Question

I need to find the left and right side limit of the function stated below, but I'm having a hard time simplifying the equation. The question is:

$$f:\mathbb R \to \mathbb R$$$$f(x)=\begin{cases} \displaystyle \frac{\sqrt{|1+x|}-1}{x} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$

$$\lim_{x \rightarrow 0^+} \frac{\sqrt{(|1+x|)}-1}{x} = \lim_{x \rightarrow 0^+} \frac{-\sqrt{(1+x)}-1}{x}$$

Please guide me on how to proceed further, as I'm kind of lost.

Answer 1

Check your signs. Since $x + 1 > 0$ near $0$, both sided limits are given by

$$ \lim_{x\to 0} \frac{\sqrt{1+x}-1}{x} = \lim_{x\to 0} \frac{(1+x)-1}{x(\sqrt{1+x}+1)} = \lim_{x\to 0}\frac{1}{\sqrt{1+x}+1} = \frac{1}{2} $$

Answer 2

Dylan's answer is likely the best approach here, but a different way might be to observe $$\lim_{x \to 0} \frac{\sqrt{|1+x|} - 1}{x} = \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}$$ and you can recognise this limit as the definition of the derivative of $\sqrt{1+x}$ at $x=0$. We know that $\frac{d}{dx} \sqrt{1+x} = \frac{1}{2} (1+x)^{1/2}$, which is $0$ at $x=0$, so both limits are equal to $\frac{1}{2}$.