# Prove limits using epsilon-delta proof

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

• the thing to be made small is $\left| 1 - \frac{x}{2x+1} \right| .$ First combine the difference into a single fraction Commented Sep 11, 2022 at 3:26
• You have a legitimate attempt but you placed it in a link. I suspect that if you formatted your attempt in your post, it would be better received. You really don't deserve the down-vote. Commented Sep 11, 2022 at 3:39
• Thank you! I'm new to this @Osmium Commented Sep 11, 2022 at 4:01
• @diamondapple123 use the buttons above the question container. Using them you can add links, images, blocks, and bold, and underline texts, and many more things. Commented Sep 11, 2022 at 4:04
• okay will take note. thank you!! @Osmium Commented Sep 11, 2022 at 4:09

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|

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| 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 $$-1-0.25 with $$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.