limit please help me ? first exam tomorrow .?

Question

find the limit when x approaches to infinity ?
(ln (2 + e^(3x))) / (in ( 1 + e ^x))

Answer 1

Put = e^x. Then, e^(3x) = t^3 and t -->oo as x--> oo.

For every t, we have

ln(2 + t^3)/ln(1 + t) = ln(t^3(1 + 2/t^3)/ln(t(1 +1/t) = [3 ln(t) + ln((1 + 2/t^3)]/[ln(t) + ln(1 +1/t)].

A t --> oo, ln(t) --> oo, 1 + 2/t^3) --> 1 and 1 + 1/t --> 1, so that ln(1 + 2/t^3) -> 0 and ln((1 + 1/t) --> 0.

Hence, [3 ln(t) + ln((1 + 2/t^3)]/[ln(t) + ln(1 +1/t)] ~ 3 ln(t)/ln(t), so that the expression tends to 3 as t --> oo.

Our limit is 3.

We could come to this same conclusion using L'Hopital. But I think my approach gives more insight.

Answer 2

Should be 3, but let me check one more time