Definition of a limit (1 Viewer)

leehuan · Jun 8, 2016

(Sorry, post remade because I made a massive error)

How would you prove that this is false from the definition? (i.e. proof by contradiction or something similar)

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

$$So the definition says $\forall \epsilon > 0 \exists M $ such that $x>M \implies |f(x)-L|< \epsilon$
$$For $x \ge 1$

$\begin{align*}\left| \frac{1}{x^2} - 1 \right| & < \epsilon \\ \iff -\frac{1}{x^2}+1 &< \epsilon \\ \iff 1-\epsilon &< \frac{1}{x^2} \\ \iff x &< \frac{1}{\sqrt{1-\epsilon}} \end{align*}$

So how should I contradict it? It can't be enough just to say the inequality is facing the wrong way can it...?

InteGrand · Jun 8, 2016

leehuan said:
(Sorry, post remade because I made a massive error)

How would you prove that this is false from the definition? (i.e. proof by contradiction or something similar)

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

$$So the definition says $\forall \epsilon > 0 \exists M $ such that $x>M \implies |f(x)-L|< \epsilon$
$$For $x \ge 1$

$\begin{align*}\left| \frac{1}{x^2} - 1 \right| & < \epsilon \\ \iff -\frac{1}{x^2}+1 &< \epsilon \\ \iff 1-\epsilon &< \frac{1}{x^2} \\ \iff x &< \frac{1}{\sqrt{1-\epsilon}} \end{align*}$

So how should I contradict it? It can't be enough just to say the inequality is facing the wrong way can it...?

$\noindent To show that $\lim _{x\to \infty}f(x) \neq L$, we would need to show that there exists an $\varepsilon >0$ such that for any $M > 0$ there will exist an $x_0>M$ such that $|f(x_0)-L|\geq \varepsilon$. (If you know a bit about formal logic, all I've done here is take the negation of the logical statement definition of a limit to infinity.)$

InteGrand · Jun 8, 2016

leehuan said:
(Sorry, post remade because I made a massive error)

How would you prove that this is false from the definition? (i.e. proof by contradiction or something similar)

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

$$So the definition says $\forall \epsilon > 0 \exists M $ such that $x>M \implies |f(x)-L|< \epsilon$
$$For $x \ge 1$

$\begin{align*}\left| \frac{1}{x^2} - 1 \right| & < \epsilon \\ \iff -\frac{1}{x^2}+1 &< \epsilon \\ \iff 1-\epsilon &< \frac{1}{x^2} \\ \iff x &< \frac{1}{\sqrt{1-\epsilon}} \end{align*}$

So how should I contradict it? It can't be enough just to say the inequality is facing the wrong way can it...?

$\noindent So intuitively we need to show that there is some `tolerance' level $\varepsilon >0$ for which we can't get our function to go and stay within that tolerance level of the claimed limit of $1$ here. (Since if we can show this, it means our function can't get and stay arbitrarily close to $1$, meaning the limit to infinity isn't 1.)$

$\noindent For example, take $\varepsilon = \frac{1}{2}$. Then for any $M>0$, take $x_0$ to be some number greater than $\max \left(3,M\right)$, so then $x_0 >M$ and $x_0 >3$, so $0<\frac{1}{x_0 ^2}<\frac{1}{9}\Rightarrow \left |\frac{1}{x_0 ^2}-1\right|\geq \frac{1}{2}=\varepsilon$.$

$\noindent So there exists an $\varepsilon >0$, namely $\frac{1}{2}$ (or many other choices) such that for any $M>0$, there exists some $x_0>M$ with $\left |\frac{1}{x_0 ^2}-1\right|\geq \varepsilon$. Hence the limit is not $1$.$

Definition of a limit (1 Viewer)

leehuan

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InteGrand

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InteGrand

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