Prove by limit definition the following: $$\begin{equation*} \lim_{x \rightarrow -1^{+}} \frac{1}{x^2 - 1}= -\infty \end{equation*}$$
Domain = $\mathbb{R}$ - {-1,+1}, so each neighborhood will contain -1 & +1, what shall I do?
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We wish to show that for every $K\gt0$ there exists a $\delta \gt0$ such that $f(x)=\dfrac{1}{x^2-1}\lt-K$ for every $x \in \mathbb{R-\{-1,1\}}$ with $0\lt|x-(-1)|=|x+1|\lt\delta$.
For $\delta=\dfrac{1}{K(x-1)}$ we have $0\lt|x+1|\lt \dfrac{1}{K(x-1)}\Rightarrow|x+1|(x-1)\lt\dfrac1K$ or equivalently $\dfrac{1}{(x+1)(x-1)}\lt-K$
MathematicianByMistake
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How you choosed delta? – Nov 03 '17 at 22:35
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@Idonotknow By looking at the last inequality-the one we want to show-and working kind off "in the reverse". – MathematicianByMistake Nov 03 '17 at 22:37
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could yo write this in details please? – Nov 03 '17 at 22:43
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@Idonotknow More detail? :) The concept is that we need to show that for no matter how small a number we chose ($-K$-so no matter how big $K$ is), the function will be even smaller close to the value $-1^{+}$. So for any such $K$ we show that we can chose a $\delta(K)$.$(\delta $ depends on $K$ $)$ The particular choice of $\delta$ is dependent on the form of the function. To find what is fitting consider the limit $\lim |f(x)-a|$ if the problem wants you to show that $f(x)\rightarrow a$. This will tell you what the $\delta$ should be. – MathematicianByMistake Nov 03 '17 at 22:50
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I do not understand how the inequality after equivalently come from the one before it ..... could you explain please? – Nov 03 '17 at 22:53
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why did not you wrote |x-1|? – Nov 03 '17 at 23:00
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@Idonotknow I should have removed the absolute value. – MathematicianByMistake Nov 03 '17 at 23:09
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delta can not contain a variable >>>>> I think you have to revise your solution – Nov 03 '17 at 23:15
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Let us continue this discussion in chat. – MathematicianByMistake Nov 03 '17 at 23:23
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Check this- https://math.stackexchange.com/questions/305793/how-exactly-cant-delta-depend-on-x-in-the-definition-of-uniform-continuity What you say is true only for uniform continuity. – MathematicianByMistake Nov 03 '17 at 23:27