Find $\lim _{x\to 0}\frac {1}{e^x +1}$.
I claim that $\lim _{x\to 0}\frac {1}{e^x +1} = 1/2$.
Proof : Let $0< \varepsilon < 1.$ Let $\delta = \min\{\ln(1+\varepsilon), \ln(\frac {1}{1-\varepsilon})\}$
Then, if $|x|< \delta,$ we can set the bound when $x \ge 0$, $e^x < e^\delta$. When $x < 0, e^x >e^{-\delta}$.
Then, $|\frac {1}{e^x +1} - \frac 12| = |\frac {1 - e^x}{2(e^x + 1)}| < |\frac {1-e^x}{2}| $.
First, consider $x\ge 0$. $|\frac {1-e^x}{2}|= \frac {e^x -1}{2} <\frac {e^\delta - 1}{2} = \varepsilon/2. $
Second, consider $x <0,$ $|\frac {1-e^x}{2}| = \frac {1-e^x}{2} < \frac {1-e^{-\delta}}{2} = \varepsilon /2 $.
Is this correct? I am not sure if I am allowed to consider two separate cases.
This is slightly more elegant:
You need to prove that for every $\varepsilon>0$ there exists $\delta$ such that:
$$\left|x\right|<\delta\implies\left|\frac{1}{1+e^x}-\frac12\right|<\varepsilon\qquad (1)$$
The statement:
$$\left|\frac{1}{1+e^x}-\frac12\right|<\varepsilon$$
...is equivalent to:
$$\frac12-\varepsilon<\frac{1}{1+e^x}<\frac12+\varepsilon$$
Suppose that $0<\epsilon<\frac12$. You can transform the last expression in the following way:
$$\frac{1-2\varepsilon}{2}<\frac{1}{1+e^x}<\frac{1+2\varepsilon}{2}$$
$$\frac{2}{1-2\varepsilon}>1+e^x>\frac{2}{1+2\varepsilon}$$
$$\frac{2}{1-2\varepsilon}-1>e^x>\frac{2}{1+2\varepsilon}-1$$
$$\frac{1+2\varepsilon}{1-2\varepsilon}>e^x>\frac{1-2\varepsilon}{1+2\varepsilon}$$
$$\ln\frac{1+2\varepsilon}{1-2\varepsilon}>x>\ln\frac{1-2\varepsilon}{1+2\varepsilon}$$
If you introduce:
$$\delta(\varepsilon)=\ln\frac{1+2\varepsilon}{1-2\varepsilon}\qquad(2)$$
...you get:
$$\delta>x>-\delta$$
...or:
$$|x|<\delta$$
So for every $\varepsilon<\frac12$ there exists $\delta(\varepsilon)$ defined by (2) such that relation (1) holds.
For $\varepsilon\ge\frac12$ you can pick $\delta$ calculated for any $\varepsilon<\frac12$.
EDIT: Say that you want to calculate $\delta$ for some $\varepsilon_1 \ge 0.5$ so that (1) holds. Obviously, you cannot use expression (2). Now pick any $\varepsilon_2 \lt \frac12$ and calculate $\delta(\varepsilon_2)$. In this case:
$$\left|x\right|<\delta(\varepsilon_2)\implies\left|\frac{1}{1+e^x}-\frac12\right|<\varepsilon_2<\varepsilon_1$$
In other words, for every $\varepsilon_1\ge0.5$ you are still able to find $\delta$ so that (1) still holds. You can use any $\delta$ calculated for smaller values of $\varepsilon$.
Please have in mind that, when using $\varepsilon-\delta$ definition to prove limits, the challenge is usually to prove (1) for sufficiently small values of $\varepsilon$. If you have a solution that works for "small" values of $\varepsilon$ (say, for $\varepsilon<0.5$), that solution also works for all "bigger" values ($\varepsilon\ge0.5$).