Algebra Word Problem Solutions

Recent questions in Algebra
Algebra IAnswered question
ajumbaretu ajumbaretu 2022-11-12

Find two consecutive integers whose sum is 73

Algebra IIAnswered question
bucstar11n0h bucstar11n0h 2022-11-12

Could somebody validate my proof regarding the limit of ln ( x n ) when, x n tends to a?
So, let me cearly state the problem:
Let ( x n ) be a convergent sequence, with: x n > 0, n, n natural number, and x n a, with a > 0. Then ln x n ln a
Here is my idea for a proof:
Our goal is to proof that there ϵ > 0 there is some n ϵ , such that n n ϵ , we have that | ln x n ln a | < ϵ
So here is what I did. First:
| ln x n ln a | = | ln x n a |
Then, beacause ln x < x, x > 0, it easily follows that   | ln x | < x, x > 0. Applying this, we have that:
| ln x n ln a | = | ln x n a | < x n a = x n a + a a = x n a a + 1
Now, we use the basic property of the absolute value: x n a | x n a | , that gives us:
| ln x n ln a | < x n a a + 1 | x n a | a + 1
Now, we use the fact that x n a. So, ϵ > 0 there is some n ϵ , such that n n ϵ , we have that | x n a | < ϵ. We choose an epsilon that takes the form a ( ϵ 0 1 ). This choice is possible for any ϵ 0
Now, we have managed to obtain that:
| ln x n ln a | < a ( ϵ 0 1 ) a + 1 = ϵ 0 , n n ϵ
Since ϵ 0 > 0 we can find an n ϵ , such that n n ϵ , the above inequality is staisfied, our claim is proved.
So, can you please tell me if my proof si correct? I've tried to find a proof, only for the limit of sequences! This problem has been on my nerves for s while. Also, probably there is some simpler way to do it, but I couldn't find it.

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