Give a δ-ε proof to prove that the limit as x approaches 3 for 5x-4=11?
i am really stuck on this and would really appreciate it.
i will choose best answer for the first correct worked solution
thanks alot!
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Verified answer
Let ε > 0. Notice that we want
|5x - 4 - 11| < ε ->
|5x - 15| < ε ->
5|x - 3| < ε ->
|x - 3| < ε/5
So pick δ = ε/5. Then
0 < |x - 3| < δ ->
|x - 3| < ε/5 ->
5|x - 3| < ε ->
|5x - 15| < ε ->
|5x - 4 - 11| < ε.