Prove the limit $\lim_{x\rightarrow-1^{+}} = \frac{1}{x^{^{2}} -1}$ exists.

Question

For each of the following, use definitions (rather than limit theorems) to prove that the limit exists. Identify the limit in each case.

(c) $\lim_{x\rightarrow-1^{+}} = \frac{1}{x^{^{2}} -1}$

Proof: By definition the function f(x) is said to converge to infinity as x → a if and only if there is an open interval I containing and given a real M, there is an δ > 0 such that 0 < |x - a| < δ implies f(x) > M, in which case f(x) approaches infinity as x → a.

Let L = infinity, and suppose ε > 0. And suppose M > 0. Then there is an δ > 0 such that |x - (- 1) | < ε . Then choose M = Can someone please help me prove the limit exists. I don't know how to continue. Please, I would really appreciate it. Thank you.

Answer 1

For any $\;M\in\Bbb R^+\;$ and $\;x>-1\;$ (but very close to $\;-1\;$)

$$\frac1{|x^2-1|}>M\iff x+1<\frac1{M|x-1|}<\frac1{2M}\implies$$

since we can make sure that $\;|x-1|>\frac32\iff\frac1{|x-1|}<\frac23\;$ , so we can choose $\;\delta_M:=\frac2{3M}\;$ , and thus:

$$x+1<\delta_M\implies\left|\frac1{x^2-1}\right|>M$$

and the above proves

$$\lim_{x\to -1^+}\frac1{x^2-1}=-\infty$$

since $\;x<-1\implies x^2-1<0\;$

Answer 2

The answer by Timbuc can be used to take care of the required estimates, so the material below consists only of preliminary comments.

You are asked to prove that $\lim_{x\to -1^+} \frac{1}{x^2-1}$ exists. Draw a number line, with $x=-1$ on it, also probably $x=0$, and maybe $x=1$.

First we try to find out what's going on. We are approaching $-1$ from the right. Imagine $x$ a tiny bit bigger than $-1$, like $-0.99999$. What can we say about $\frac{1}{x^2-1}$? The number $x^2$ is then close to $1$, but less than $1$. So $x^2-1$ is close to $0$, but negative. It follows that $\frac{1}{x^2-1}$ is "large negative."

The way some people use limit language, that means the limit does not exist. Other people say that the limit "is" $-\infty$. It looks as if in your course "infinite" limits are allowed. We want to prove that $$\lim_{x\to -1^+}\frac{1}{x^2-1}=-\infty.$$ So we want to prove that for any positive $M$, there is a $\delta\gt 0$ such that if $-1\lt x\lt -1+\delta$, then $\frac{1}{x^2-1}\lt -M$. (The $L$ and $\epsilon$ stuff in the post should be deleted.)

The estimates by Timbuc can now be used.