Rate of growth of linear and logarithmic functions. I have the following limit: lim_(x -> +oo) (log(e^(2x+1)+x))/(x)=2

Macy Villanueva

Macy Villanueva

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2022-08-20

I have the following limit:
lim x + log ( e 2 x + 1 + x ) x = 2
I know for sure that
lim x + log x x = 0
Because x grows a lot faster than log x . Then why wouldn't the first limit be equal to 0 as well?

Answer & Explanation

Leon Clark

Leon Clark

Beginner2022-08-21Added 9 answers

because it has that e 2 x + 1 in the numerator it is making all the difference
lim x + log ( e 2 x + 1 ) x lim x + log ( e 2 x + 1 + x ) x
lim x + ( 2 x + 1 ) x lim x + log ( e 2 x + 1 + x ) x
2 lim x + log ( e 2 x + 1 + x ) x
find the limit using Lhopital's rule
lim x + log ( e 2 x + 1 + x ) x = lim x 0 e 2 x + 1 2 + 1 e 2 x + 1 + x
lim x + log ( e 2 x + 1 + x ) x = lim x 0 2 + 1 e 2 x + 1 1 + x e 2 x + 1 = 2

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