Using $\epsilon$-$\delta$ definition to prove that $\lim_{x\to-2}\frac{x-1}{x+1}=3$.

Question

Prove $$ \lim_{x\rightarrow -2} \frac{x-1}{x+1}=3. $$

I am stuck on this question please help me with it! I really do not know how to start effectively. I am new to this very precise, abstract form of thinking.

Answer 1

I am only going to answer this because I know how tricky these problems are for beginners (took me a long time to understand how to do them as well--I understand that even beginning effectively can be rather difficult). Thus, in the future, at least show some work or ideas on how you thought about going about it. Otherwise, you cannot count on getting good answers. That being said consider the following: (1) In the first part, I'll go through the process of guessing a value for $\delta$ [this is really an art form], and (2) I'll show you that this $\delta$ does, indeed, work.

Part 1 (Guessing a value for $\delta$): Let $\epsilon>0$ be given. We have to find a number $\delta>0$ such that $$ \left|\frac{x-1}{x+1}-3\right|=\left|\frac{x-1-3x-3}{x+1}\right|=\left|\frac{-2x-4}{x+1}\right|=\left|\frac{-2(x+2)}{x+1}\right|=\frac{2}{|x+1|}\cdot|x+2|<\epsilon $$ whenever $0<|x+2|<\delta$. We find a positive constant $C$ such that $$ \frac{2}{|x+1|}<C\Rightarrow \frac{2}{|x+1|}\cdot|x+2| < C|x+2|, $$ and we can make $C|x+2|<\epsilon$ by taking $|x+2|<\frac{\epsilon}{C}=\delta$. We restrict $x$ (note: this is where part of the "art form" comes into play) to lie in the interval $|x+2|<\frac{1}{2}$, whereby we note that $$ |x+2|<\frac{1}{2}\Longleftrightarrow -\frac{3}{2}<x+1<-\frac{1}{2}\Longleftrightarrow -\frac{2}{3}>\frac{1}{x+1}>-2\Longleftrightarrow -\frac{4}{3}>\frac{2}{x+1}>-4. $$ Thus, certainly $$ \frac{2}{x+1}>-4 \Longrightarrow 4>\frac{2}{|x+1|}>-4. $$ So $C=4$ is suitable. Thus, we should choose $\delta=\min\{\frac{1}{2},\frac{\epsilon}{4}\}$. Now we need to check if this $\delta$ really works.

Part 2 (Showing that $\delta$ works): Given $\epsilon>0$, we let $\delta=\min\{\frac{1}{2},\frac{\epsilon}{4}\}$. If $0<|x+2|<\delta$, then $\frac{2}{|x+1|}<4$, as we saw above in Part 1. Also, $|x+2|<4\epsilon$, so $$ \left|\frac{x-1}{x+1}-3\right|=\frac{2}{|x+1|}\cdot|x+2|<4\cdot\frac{\epsilon}{4}=\epsilon. $$

This shows that $$ \lim_{x\to -2}\frac{x-1}{x+1}=3.\quad\blacksquare $$