How to find the $\delta$ for some $\varepsilon$ such that: $|x-0| < \delta \Rightarrow |g(x)-g(0)|<\varepsilon$

Question

$g(x)$ is defined as:

$$g(x)=\frac{\sqrt{x+1}-1}{x};\ x\in [-1,0)\ \cup\ (0,\infty)$$

$$g(x)=\frac{1}{2};\ x=0$$

How can I find the $\delta$ for some $\varepsilon$ such that: $|x-0| < \delta \Rightarrow |g(x)-g(0)|<\varepsilon$.

I tried a lot of things, but I seem to be completely stuck. Rationalizing the numerator doesn't help, because x cancels out.

Answer 1

Observe that \begin{align*} |g(x) - g(0)| &= \left| \dfrac{\sqrt{x+1} - 1}{x} -\dfrac{1}{2} \right| \\ &= \left| \dfrac{1}{\sqrt{x+1} + 1} - \dfrac{1}{2} \right| \\ &= \left| \dfrac{1 - \sqrt{x+1}}{2(\sqrt{x+1} + 1)} \right| \\ &= \left| \dfrac{-x}{2(\sqrt{x+1} + 1)^2} \right| \\ &= \dfrac{|x|}{2(\sqrt{x+1} + 1)^2} \\ &< \dfrac{\delta}{2} \end{align*}

So we can choose $\delta < 2 \epsilon$.

However, the domain of $g$ is $[-1,\infty)$. If $\delta > 1$, any $c$ in $(-\delta,-1)$ such as $c = -\dfrac{\delta + 1}{2}$ satisfies $|c| < \delta$ but $g(c)$ does not exist, thus $|g(c) - g(0)| < \epsilon$ is false.

So one of the answer is $\delta < \min\{ 1, 2 \epsilon \}$.

Answer 2

Hint: $$|g(x) - g(0)| = \left|\frac{\sqrt{x+1}-1}{x} - \frac{1}{2}\right| = \left|\frac{1}{\sqrt{x+1}+1} - \frac{1}{2} \right| < \varepsilon$$ Solving the inequality for $x$ you get $$\varepsilon - \frac{1}{2} < \frac{1}{\sqrt{x+1}+1} < \varepsilon + \frac 12 \\ \frac{1}{\varepsilon + 1/2} < \sqrt{x+1} + 1 < \frac{1}{\varepsilon - 1/2} \\ \left(\frac{1/2 - \varepsilon}{\varepsilon + 1/2}\right)^2 -1 < x < \left(\frac{3/2-\varepsilon}{\varepsilon - 1/2} \right)^2 - 1$$