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I am wondering if there is a way to prove the following functional limit, only by definition.

$\lim_{x\rightarrow 3}(1/x)=1/3$

Currently, my proof involves proving separately that lim 3-x goes to 0 and lim 3x goes to 9 as x goes to 3. Then, I use the Algebraic Theorem for Functional limits to prove to original proposition. Again, the question is simply, how might I do this without the Theorem? Is that possible?

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  • $\begingroup$ It is always possible for such simple cases. $\endgroup$ – Crostul Mar 22 '16 at 22:20
  • $\begingroup$ Do you have a hint? $\endgroup$ – user322548 Mar 22 '16 at 22:22
  • $\begingroup$ $$\left| \frac{1}{x} - \frac{1}{3} \right| = \frac{|x-3|}{3|x|} = \frac{\mbox{small}}{\mbox{bounded away from $0$}} = \mbox{small}$$ now pick $\delta = \min \{ \varepsilon , 1\}$. $\endgroup$ – Crostul Mar 22 '16 at 22:25
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$\forall \epsilon >0, \exists \delta > 0$ such that $|x-3|<\delta \implies |1/x-1/3|<\epsilon$

$|x-3|<\delta \implies 3-\delta <x<3+\delta$

let $\delta = \min(1,6\epsilon)$

x > 2

$|1/x-1/3|<\delta/6\leq\epsilon$

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  • $\begingroup$ Could you elaborate on how it is that |x-3|<d implies |1/x - 1/3|<epsilon? The connection to the choice of delta is not all that clear to me. $\endgroup$ – user322548 Mar 23 '16 at 0:20
  • $\begingroup$ |1/x - 1/3| = |(x-3)/3x|, |x-3|<delta ==> x>2 and |1/x - 1/3| < delta / 6 $\endgroup$ – Doug M Mar 23 '16 at 0:32
  • $\begingroup$ Ah, I see. Thank you. $\endgroup$ – user322548 Mar 23 '16 at 0:54

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