Epsilon-delta proof verification (only finding good delta)

Question

$\varepsilon-\delta$ proof of $$ \lim_{x \to -\frac{1}{2}} \frac{1}{x+1} $$

Here's how I found $\delta$: We got to somehow bound $\delta$ with $\epsilon$, so let's find $|x + \frac{1}{2}|$ in $|\frac{1}{x+1} - 2|$. We have: $$ \left |\frac{1}{x+1} - 2 \right | = \left |\frac{1}{x+1} - \frac{2(x+1)}{x+1} \right | = \left |\frac{1 - 2x - 2}{x+1} \right | = \left |\frac{-2(x + \frac{1}{2})}{x+1} \right | = \frac{2|x + \frac{1}{2}|}{|x+1|} $$

So let's see what's happening when $|x + \frac{1}{2}| < 1$. We have $ -\frac{1}{2} < x < \frac{3}{2} $, so upper bound for it is $\frac{3}{2}$. We can bound $|x + 1|$ using this information - it wouldn't be smaller than $\frac{1}{2}$. Thus $$ \frac{2|x + \frac{1}{2}|}{|x+1|} < \frac{2}{|x+1|} < \frac{2}{\frac{1}{2}} = 4 $$ Therefore our delta is $\delta = \min \{ 1, \frac{\varepsilon}{4} \}$

Is this a good delta?

Answer 1

If you pick$|x+0.5|<1$, we have $$-1 < x+0.5< 1$$

$$-1.5<x<0.5$$

then $$-0.5 < x+1<1.5$$

then $|x+1|$ might take value $0$.

Let's make the $\delta$ smaller.

We want $|x+0.5|< \delta$, $$-\delta < x+0.5 < \delta$$

$$-0.5-\delta < x < -0.5+\delta$$

$$0.5-\delta < x+1<0.5+\delta$$

Let $\delta < 0.25$, then $|x+1|>\frac14$.

$$\frac{2|x+0.5|}{|x+1|}\le 8\delta$$

Answer 2

Hint :

By the Triagnle Inequality, one can yield :

$$\bigg|x + \frac{1}{2}\bigg| = \Bigg|(x + 1) + \bigg(-\frac{1}{2}\bigg)\Bigg|\leq |x+1| + 1/2$$

Thus :

$$\Bigg|\frac{1}{x+1}-2\Bigg| \leq \frac{2|x+1|+1}{|x+1|}=2+\frac{1}{|x+1|}$$

So, what can be concluded ?