$\lim_ {x\rightarrow−2} \frac{1}{x + 1} = −1$

Question

Been repeatedly stumped by this limits, epsilon-delta question: $\lim_ {x\rightarrow−2} \frac{1}{x + 1} = −1$. The answer in the back of Anton Transcendentals is: Min$(\frac{1}{2}, \frac{\epsilon}{2})$ I have tried manipulating it every which way, but cannot get my numbers to come out right, any help greatly appreciated.

Answer 1

If we set $f(x)=\frac{1}{x+1}$, then $$f(-2)=\frac{1}{-2+1}=-1\in\mathbb{R}$$

Hence, $$\lim_{x\to-2}f(x)=f(-2)=-1$$

No need to manipulate the numbers .

Answer 2

First, try to find an upper $B$ bound on $\left\lvert\frac{f(x)-L}{x-a}\right\rvert$ on some $\alpha$-neighborhood $(a-\alpha,a+\alpha)$ of $a$.

\begin{eqnarray} \left\lvert\frac{f(x)-L}{x-a}\right\rvert&=&\left\lvert\frac{\frac{1}{x+1}-(-1)}{x-(-2)}\right\rvert\\ &=&\left\lvert\frac{x+2}{(x+1)(x+2)}\right\rvert\\ &=&\left\lvert\frac{1}{x+1}\right\rvert& \end{eqnarray}

Now we want to find an $\alpha$-neighborhood $(-2-\alpha,-2+\alpha)$. Usually, one chooses something simple, such as $\alpha=1$, but this time we will need something smaller such as $\alpha=\frac{1}{2}$

$$ -\frac{5}{2}<x<-\frac{3}{2}$$

and look for an upper bound on $B$ on$\left\lvert\frac{1}{x+1}\right\rvert$ on that interval.

\begin{eqnarray} -\frac{5}{2}&<&x<-\frac{3}{2}\\ -\frac{3}{2}&<&x+1<-\frac{1}{2}\\ \frac{1}{2}&<&-(x+1)<\frac{3}{2}\\ \frac{2}{3}&<&\left\lvert\frac{1}{x+1}\right\rvert<2 \end{eqnarray}

So $B=2$ is an upper bound on $\left\lvert\dfrac{\frac{1}{x+1}-(-1)}{x-(-2)}\right\rvert\ \text{when} -\frac{5}{2}<x<-\frac{3}{2}$

Now we are done with the preliminary investigations and are ready to present the actual proof that

$$ \lim_ {x\rightarrow−2} \frac{1}{x + 1} = -1 $$

by using $\delta=\min\{\frac{1}{2},\frac{\epsilon}{B}\}$.

Proof: Let $\epsilon>0$. Let $\delta=\min\{\frac{1}{2},\frac{\epsilon}{2}\}$ and let $x\in(-2-\delta,2)\cup(-2+\delta)$. Then two things follow as a result:

1. $\left\lvert\frac{\frac{1}{x+1}-(-1)}{x-(-2)}\right\rvert<B$
2. $|x-(-2)|<\frac{\epsilon}{B}$

Thus, multiplying the two inequalities gives

$$ \left\lvert\frac{1}{x-1}-(-1) \right\rvert<\epsilon $$

Therefore,

$$ \lim_ {x\rightarrow−2} \frac{1}{x + 1} = -1 $$

ADDENDUM: As an exercise, try using $\alpha=1$ rather than $\alpha=\frac{1}{2}$ and see why that fails for this particular limit.

Answer 3

$f(x)=|\dfrac{1}{x+1}-(-1)|=|\dfrac{x-(-2)}{x+1}|.$

For $|x-(-2)| <1/2,$ we have

$-1/2 < x+2<1/2$, $-3/2 <x+1<-1/2$, i.e.

$|x+1| >1/2$.

Let $\epsilon >0$ be given.

Choose $\delta =\min(1/2,\epsilon/2)$.

Then $|x-(-2)|<\delta$ implies

$f(x)=\dfrac{|x+2|}{|x+1|} < \dfrac{\delta}{1/2}= 2\delta <\epsilon.$