Proving a limit doesn't exist

Question

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?

Answer 1

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.

Answer 2

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 $x\to 1$?

Answer 3

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 $x_n = 1-1/n ?$

Answer 4

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 $\mathbb{R}\cup{\{-\infty,+\infty \}}$ $)$

Answer 5

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.