Arnold F. answered 03/28/16
Tutor
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College Professor & Expert Tutor In Statistics and Calculus
What you want to show is that for every ε>0 there exists a number k such that when x>k then |1/ln(x) - 0| <ε.
Since x goes to infinity without loss of generality we can remove the absolute value sign.
We want to show (1/ln(x)) < ε for x > k.
This gives 1/ε < ln(x). Solve for x and you are one step away from being done.
Let me know if you have any questions.
Alsina S.
03/28/16