I'm not very good with analysis (I never studied it) but because of my "work" on other topics of mathematics I came to this problem. lim_(k->+oo) log_k(k^a+k^b)=max(a,b) I'm sure that this is really "reasonable" because I tryed to graph it for really huge values of k ... but "reasonable" is not enough in mathematics: I'd like to prve this. So I'm courious to know how one should go to prove it in a formal way ... if it is true (I hope). PS: I know nothing about Limits and their rules

PoentWeptgj

PoentWeptgj

Answered question

2022-07-14

What is the limit of log k ( k a + k b ) for k + ?
I'm not very good with analysis (I never studied it) but because of my "work" on other topics of mathematics I came to this problem.
lim k + log k ( k a + k b ) = max ( a , b )
I'm sure that this is really "reasonable" because I tryed to graph it for really huge values of k ... but "reasonable" is not enough in mathematics: I'd like to prve this. So I'm courious to know how one should go to prove it in a formal way ... if it is true (I hope).
PS: I know nothing about Limits and their rules

Answer & Explanation

juicilysv

juicilysv

Beginner2022-07-15Added 17 answers

Use the functional equation:
log ( k a + k b ) = a + log ( 1 + k b a )
When a > b the log term goes to 0, if b a it goes to b a and b a + a = b
Lexi Mcneil

Lexi Mcneil

Beginner2022-07-16Added 2 answers

Without loss if generality assume a b then:
log k ( k a + k b ) = a + log k ( 1 + k b a ) = a + ln ( 1 + k b a ) ln ( k )
where I changed the base of the logarithm to get the second equality. Notice that if a > b then 1 + k b a 1 and ln ( k ) both as k tends to . If a = b then 1 + k b a = 2 but ln ( k ) as k . Thus, the above tends to a = max { a , b }

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?