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Let's say we have this limit: $$\lim\limits_{x\to \infty} \frac{1}{x}$$ which is clearly $$\lim\limits_{x\to \infty} \frac{1}{x} = 0.$$ From there, to prove it we should: $$\left\lvert \frac{1}{x} - 0 \right\rvert < \epsilon$$ (with $\epsilon > 0$ and small).

To solve that inequality we should deal with a system of: $$\begin{align*} \frac{1}{x} &\lt \epsilon&&\text{(for }\frac{1}{x} \gt 0\text{)}\\ \frac{1}{x} &\gt -\epsilon&&\text{(for }\frac{1}{x}\lt 0\text{)} \end{align*}$$ Then from the () we have that the first inequality is for $x < 0$ and the second is for $x > 0$.

Is this right?

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1 Answer 1

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As you are dealing with a limit for $x \rightarrow +\infty$ we can assume that $x > 0$ and so we only need to consider $\frac{1}{x} < \epsilon$ which is equivalent to $x > \frac{1}{\epsilon}$.

So for any $\epsilon > 0$, for all $x > \frac{1}{\epsilon} (> 0)$ we have that $\lvert \frac{1}{x} - 0 \rvert < \epsilon$, as required.

[edit] The limit for $x \rightarrow \infty$ must mean (in our case) that $$\forall \epsilon > 0, \exists K>0: \forall x \in \mathbb{R}: ( \lvert x \rvert > K \rightarrow \lvert \frac{1}{x} - 0 \rvert < \epsilon)$$

which is clear when we take $K = \frac{1}{\epsilon}$.

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I didn't specified that $x \rightarrow +\infty$. $x \rightarrow \infty$ could be considered as $x \rightarrow +\infty$ and/or $x \rightarrow -\infty$ –  user9209 Feb 12 '12 at 16:11
    
@CharliePigarelli: In that case I would write $\lvert x\rvert \to \infty$. This notation can be used in any normed vector space –  kahen Feb 12 '12 at 17:17
    
@CharliePigarelli: That is very non-standard notation. $\infty$ and $+\infty$ are usually considered to be the same thing in calculus/real analysis. –  Arturo Magidin Feb 12 '12 at 22:40

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