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If $\displaystyle \lim_{n\to\infty} a_n=L$ then prove using the limit definition that: $\displaystyle \lim_{n\to\infty} -2a_n=-2L$.

From the given and the definition we know that: $L-\epsilon<a_n<L+\epsilon$ multiply that by $-2$ and we get: $2\epsilon-2L>-2a_n>-2L-2\epsilon\Rightarrow |-2a_n+2L|<2\epsilon$ which concludes that: $\displaystyle \lim_{n\to\infty} -2a_n=-2L$.

I feel like I was cheating by doing that multiplication by $-2$, is it alright? it's still true for $2\epsilon$ right?

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  • $\begingroup$ Your proof is almost right. You can take $\varepsilon$ small enough although 2 is multiplied. The only defect of your proof is absence of mentioning a lower bound $N$ of $n$ (which depends on $\varepsilon$.) $\endgroup$ – Hanul Jeon Dec 14 '14 at 14:38
  • $\begingroup$ @tetori so what's left is just mentioning there exists $N$ such that it's true for all $n>N$? $\endgroup$ – shinzou Dec 14 '14 at 14:43
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    $\begingroup$ Yes, and it is guaranteed by the definition of the limit. If your proof is uncomfortable, you can consider such proof: for each $\varepsilon$, you can find $N$ such that $|a_n-L|<\varepsilon/2$ for every $n>N$ so... $\endgroup$ – Hanul Jeon Dec 14 '14 at 14:54
  • $\begingroup$ Ah right, dividing by 2 to make it more cosmetically pleasing as our professor say. Well, thank you. $\endgroup$ – shinzou Dec 14 '14 at 15:00
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The general idea of you solution is very much correct. The only thing that is missing is that we have: $L-\epsilon<a_n<L+\epsilon$, for sufficiently large $n$.

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