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Okay so I know that the answer is just C, but I don't understand how to get there. I figured out that I have to use L Hospitals rule because I have an indeterminate form of infinity over infinity. I then got (1/A + e^(Cx) * e^(Cx))/(1). I then simplified that to 1/(A+e^(Cx)) which then simplified into e^(Cx).

Can someone please guide me and explain to me the process of solving this problem? Thank you!

A and C are constants by the way!

EDIT: C and A > 0. Sorry forgot to specify that!

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  • $\begingroup$ Please learn to use LaTeX to format your questions. It's often as simple as wrapping dollar signs around your formulae. You can view the LaTeX code in any question or answer by hitting the edit button underneath to learn how it works. $\endgroup$ – Jack M Apr 25 '14 at 1:01
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We have $$\dfrac{\ln(a+e^{cx})}x = \dfrac{\ln(e^{cx}) + \ln(1+ae^{-cx})}x = c + \dfrac{\ln(1+ae^{-cx})}x$$ Assuming $c>0$, letting $x \to \infty$, the second term goes to zero and hence the limit is $c$.

If $c<0$, then the limit is $0$.

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If $A+e^{Cx}>0$, then $$\lim_{x\to\infty}\dfrac{\ln(A+e^{Cx})}{x}\stackrel{\mathrm{L'Hopital}}{=}\lim_{x\to\infty}\dfrac{Ce^{Cx}}{A+e^{Cx}}\stackrel{\mathrm{L'Hopital}}{=}\lim_{x\to\infty}\dfrac{C^2e^{Cx}}{Ce^{Cx}}=C$$ Else the limit doesn't exist in the real numbers.

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Edit: I've edited to make it crystal clear that C > 0, as it says in the question, since user141421 below disagrees with my answer.

Assuming A is a constant, it doesn't matter in the limit, since you are going to infinity.

$$\lim_{x\to\infty}\dfrac{\ln(A+e^{Cx})}{x}{=}\lim_{x\to\infty}\dfrac{\ln(e^{Cx})}{x}{=}\lim_{x\to\infty}\dfrac{{Cx}}{{x}}=\lim_{x\to\infty}C=C$$

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