Question

Precise limit definition. Prove the statement using the delta and epsilon definition of a limit.

Answer 1

\[\lim_{x \rightarrow 3}(1+\frac{ 1 }{ 3 }x)=2\]

Answer 2

So far, I have, given \[\delta < 0\] and \[\epsilon < 0\], such that if \[0 < \left| x-3 \right| < \delta \], then \[\left| (1+\frac{ 1 }{ 3 }x)-2 \right|<\epsilon\] = \[\left| \frac{ 1 }{ 3 }x-1 \right|<\epsilon \]

Answer 3

try factoring out 1/3 on the left hand side of your last inequality

Answer 4

|1/3|=1/3
and multiply both sides by 3

Answer 5

it was readable

Answer 6

you can now easily choose your delta