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So I'm having trouble proving that a limit doesn't exist, I know that you have to find an epsilon for which a delta doesn't work but I'm not sure how to do that.

For example, for the question $\lim\limits_{x\to 1} \dfrac{x}{x-1}$

How would one go about starting this?

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  • $\begingroup$ what does x/x-1 mean? x/(x-1) or (x/x)-1 ? $\endgroup$ – miracle173 Oct 14 '17 at 22:13
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    $\begingroup$ I will point out that $\frac{x}{x-1}=\frac{x-1+1}{x-1}$ by simply adding "zero" to the numerator in the form of adding and subtracting the same thing. This simplifies further then as $\frac{x-1+1}{x-1}=1+\frac{1}{x-1}$. Does this help? Have you seen the proof as to why $\lim\limits_{x\to 0}\frac{1}{x}$ doesn't exist? $\endgroup$ – JMoravitz Oct 14 '17 at 22:22
  • $\begingroup$ It's useful to be more careful about the logic. To say $L$ is the limit means $$\forall \epsilon > 0 \,\, \exists \delta > 0 \,\, \text{such that} \,\, \forall x \ne 1, \,\, \text{if} \,\, |x-1| < \delta \,\,\text{then}\,\,\bigl| \frac{x}{x-1} - L \bigr| < \epsilon $$ The negation of this, which says $L$ is not the limit, is $$\exists \epsilon > 0 \,\,\text{such that}\,\, \forall \delta > 0 \,\, \exists x \ne 1 \,\,\text{such that}\,\, |x-1| < \delta \,\,\text{and}\,\, \bigl| \frac{x}{x-1} - L \bigr| \ge \epsilon $$ $\endgroup$ – Lee Mosher Oct 14 '17 at 22:50
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Hint:

In this situation, it can be helpful to formalise the condition, and translate it in ordinary language.

The formal definition of the sentence ‘ the function $f$ has a limit at $0$ ’ is this: $$\exists \ell\in\mathbf R\;\forall \varepsilon>0\;\exists\, \delta>0\,\forall x, \;\bigl(\,|x|<\delta\implies |f(x)-\ell|<\varepsilon\,\bigr)$$ The negation of this sentence then becomes, formally: $$\forall \ell\in\mathbf R\;\exists\, \varepsilon>0\;\forall \delta>0\;\exists\, x,\;\bigl((\,|x|<\delta)\wedge (\,|f(x)-\ell|\ge\varepsilon\,)\bigr)$$ In other words, taking into account that $|a-b|$ is the distance from $a$ to $b$:

For any real number $\ell$, there will be numbers $x$ such $f(x)$ will stay at a minimal distance from $\ell$, no matter how small (close to $0$) $x$ is.

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For $1/2<x$ the term in your limit is bounded below as follows $$ \frac{\frac{1}{2}}{x-1}<\frac{x}{x-1} $$ what happens to the function on the left as $x\to 1$?

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We have $$\lim _{x{\rightarrow}1}\frac{x}{x-1}=\lim _{x{\rightarrow}1}\frac{x+1-1}{x-1}=\lim _{x{\rightarrow}1}\Big[1-\frac{1}{x-1}\Big]$$

Let $x-1=u$

This reduces the problem to figuring out the limit $$\lim _{u{\rightarrow}0}\frac{1}{u}$$

But for every $N\in \mathbb{N}:$ $N\gt0$ and for $0\lt|u|\lt\delta$ we can choose $\delta=\frac1N \Rightarrow\frac1u\gt N$

This means that $$\lim _{u{\rightarrow}0}\frac{1}{u}\rightarrow+\infty$$ So the initial limit doesn't exist in $\mathbb{R}$ $ $(but it exists in $\mathbb{R}\cup{\{-\infty,+\infty \}}$ $)$

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If we take the extended real then $x/{x-1}→∞$ as $x→1+$. Similarly $x/{x-1}→-∞$ as $x→1-$. Both left hand and right hand limits are different. So limit doesn't exist.

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Consider $(x_n)_{n \in \mathbb{N^+}} = 1+ 1/n.$

$y_n := \dfrac{x_n}{x_n -1} = $

$\dfrac{1+1/n}{1/n}.$

$y_n \gt \dfrac{1}{1/n} = n.$

$\rightarrow:$

$\lim_{n \rightarrow \infty} y_n = \infty.$

What happens if $x_n = 1-1/n ?$

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