Can't seem to prove this limit
$$\displaystyle \lim_{x \to -1}\frac{x}{2x+1}=1$$
I got this but I got stuck:
Given $\varepsilon>0$, whenever $0<|x+1|<\delta$, choose
\begin{aligned} \delta= &\left|\frac{x}{2 x+1}-1\right| \\ =&\left|\frac{x-(2 x+1)}{2 x+1}\right| \\ =&\left|\frac{-x-1}{2 x+1}\right| \\ =&\left|\frac{x+1}{2 x+1}\right| \\ =& \frac{|x+1|}{12 x+11} \end{aligned}
Image version: https://i.sstatic.net/4dsBZ.png
Prove $\displaystyle\lim_{x\to a}F(x)=L$
Sometimes you can find, given a value of $\epsilon>0$, an appropriate value of $\delta$ required by the $\epsilon,\delta$ definition by finding some $B$ such that
$$ \left|\frac{f(x)-L}{x-a}\right|<B $$
for some 'punctured' interval $(a-t,a)\cup(a,a+t)$
If so then one can let $\delta=\min\left\{t,\frac{\epsilon}{B}\right\}$.
Then it follows that $|x-a|<\frac{\epsilon}{B}$ and that $\left|\frac{f(x)-L}{x-a}\right|<B$ thus
$$ |x-a|\cdot\left|\frac{f(x)-L}{x-a}\right|=|f(x)-L|<\epsilon $$
Sometimes simply letting $t=1$ will suffice, sometimes a smaller $t$ is necessary.
$$ \lim_{x\to-1}\frac{x}{2x+1}=1 $$
In this particular example, we need a smaller $t$, such as $t=0.25$ to avoid dividing by zero, as we will see.
First let us note that
$$\left|\frac{f(x)-L}{x-a}\right|=\left|\frac{1}{2x+1}\right|$$
Let us find an upper bound on $\left|\frac{1}{2x+1}\right|$ for $-1-0.25<x<-1+0.25$ with $x\ne-1$
Then working with this inequality we will try to transform it into an inequality on $\left|\frac{1}{2x+1}\right|$.
\begin{eqnarray} -1.25&<&x<-0.75\quad x\ne-1\\ -2.5&<&2x<-1.5\\ -1.5&<&2x+1<-0.5\quad\text{Note that }2x+1\ne0\\ -2&<&\frac{1}{2x+1}<-\frac{2}{3}\\ &&\left|\frac{1}{2x+1}\right|<2 \end{eqnarray}
Let $\delta=\min\left\{0.25,\frac{\epsilon}{2}\right\}$. Then $|x-(-1)|<\frac{\epsilon}{2}$ and $\left|\frac{1}{2x+1}\right|<2$
Thus $|x+1|\cdot\left|\frac{1}{2x+1}\right|<\frac{\epsilon}{2}\cdot2=\epsilon$
That is, $\left|\frac{x+1}{2x+1}\right|<\epsilon$
NOTE: The process is actually much quicker that this. I have made it extra detailed to make it easier to understand. Once it has been practiced with several examples, it can be done quickly for limits which yield to this approach. It will not work on functions which have a vertical tangent at the limit point.
