Limit of a sum of natural logarithms As the title says I have to calculate a limit: lim_(x->0)(ln(1+x)+ln(1+2x)+...+ln(1+px))^x I've transformed the sum into one logarithm ln((1+x)(1+2x)...)etc but I don't see how it helps me further. Some hints would be great.

Angel Kline

Angel Kline

Answered question

2022-10-26

Limit of a sum of natural logarithms
As the title says I have to calculate a limit:
lim x 0 ( ln ( 1 + x ) + ln ( 1 + 2 x ) + . . . + ln ( 1 + p x ) ) x
I've transformed the sum into one logarithm ln ( ( 1 + x ) ( 1 + 2 x ) . . . ) etc but I don't see how it helps me further. Some hints would be great.

Answer & Explanation

blackcat1314xb

blackcat1314xb

Beginner2022-10-27Added 19 answers

Using
ln ( 1 + x ) 0 x
we have
( ln ( 1 + x ) + ln ( 1 + 2 x ) + + ln ( 1 + p x ) ) x 0 ( p ( p + 1 ) 2 x ) x = exp ( x ln ( p ( p + 1 ) 2 x ) ) x 0 1
podvelkaj8

podvelkaj8

Beginner2022-10-28Added 3 answers

Using sum-of-logs = log-of-product:
g ( x ) = n = 1 p ln ( 1 + n x ) = ln ( n = 1 p ( 1 + n x ) )
h ( x ) = e g ( x ) = n = 1 p ( 1 + n x )
and ( ln a ) b = b ln a
f ( x ) = g ( x ) x = ( ln h ( x ) ) x = x ln h ( x )
As x 0, each term in h ( x ) approaches 1, so h ( x ) 1, and
lim x 0 f ( x ) = lim x 0 x ln h ( x ) = 0 × ln ( 1 ) = 0 × 0 = 0.

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?