Friday 9 February 2018

calculus - Showing if $lim_{ntoinfty} a_n=L$ then $lim_{ntoinfty} -2a_n=-2L$ using definition




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: $L-\epsilon-2a_n>-2L-2\epsilon\Rightarrow |-2a_n+2L|<2\epsilon$ 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?


Answer



The general idea of you solution is very much correct. The only thing that is missing is that we have: $L-\epsilonfor sufficiently large $n$.


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