Limit of logarithmic function using l'Hospital How can I find the following limit: lim_(x-oo)(ln(1+alpha x))/(ln(ln(1+e^(beta x)))) where alpha, beta in RR^+

Melissa Walker

Melissa Walker

Answered question

2022-11-18

Limit of logarithmic function using l'Hospital
How can I find the following limit:
lim x ln ( 1 + α x ) ln ( ln ( 1 + e β x ) )
where α ,   β R +
My first guess was to use l'Hospital:
lim x ln ( 1 + α x ) ln ( ln ( 1 + e β x ) ) = lim x ln ( 1 + e β x ) ( 1 + e β x )   α ( 1 + α x )   e β x   β
But what can I do now? Is my approach correct or is there a simpler method?
Edit:
lim x ln ( 1 + e β x ) ( 1 + e β x )   α ( 1 + α x )   e β x   β = lim x ln ( 1 + e β x )   α ( 1 + α x )   β lim x ( 1 + e β x )
Applying L'Hospital a second time on the first fraction,
lim x ln ( 1 + e β x )   α ( 1 + α x )   β lim x ( 1 + e β x ) = lim x α   β   e β x ( 1 + e β x )   α   β 1 = lim x e β x 1 + e β x 1
Now let's apply L'Hospital one final time:
lim x e β x 1 + e β x 1 = lim x e β x   β e β x   β 1 = 1
Is this correct?

Answer & Explanation

martinmommy26nv8

martinmommy26nv8

Beginner2022-11-19Added 16 answers

The first application of l'Hôpital's theorem gives
lim x α 1 + α x ( β e β x / ( 1 + e β x ) log ( 1 + e β x ) ) 1 = lim x α β 1 + e β x e β x log ( 1 + e β x ) 1 + α x
Now,
lim x 1 + e β x e β x = lim x ( e β x + 1 ) = 1
so we just need to compute
lim x log ( 1 + e β x ) 1 + α x = lim x β e β x / ( 1 + e β x ) α = lim x β α e β x 1 + e β x = β α
You don't need anything new for this limit, because you have just computed the reciprocal.
Alfredo Cooley

Alfredo Cooley

Beginner2022-11-20Added 4 answers

We can proceed as follows
L = lim x log ( 1 + α x ) log ( log ( 1 + e β x ) ) = lim x log ( 1 + α x α x ) + log α x log ( log ( 1 + e β x e β x ) + β x ) = lim x y + log α x log ( z + β x )  where  y = log ( ( 1 + α x ) / α x ) , z = log ( ( 1 + e β x ) / e β x ) = lim x y + log α x log ( z + β x ) log β x + log β x = lim x y + log α x z c + log β x  where  c ( β x , β x + z )  by Mean Value Theorem = lim x y + log α x t + log β x  where  t = z / c = lim x y + log α + log x t + log β + log x = lim x y log x + log α log x + 1 t log x + log β log x + 1 = 0 + 0 + 1 0 + 0 + 1 = 1
Here we have used the fact that α > 0 , β > 0 , y > 0 , z > 0 and as x we have y 0, z 0, c and t 0

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