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Can someone help me solve this limit? $$\lim_{x\to\infty}{\log_{e}(e^x+x)+2\over x}$$ I've tried everything I can think of but always get stuck somewhere.

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closed as off-topic by choco_addicted, Brahadeesh, Claude Leibovici, Shailesh, Michael Hoppe Oct 17 '18 at 10:25

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  • $\begingroup$ If you don't realize @hHhh's nice trick below, have you tried l'Hopital's Rule? $\endgroup$ – Simon S Jun 19 '15 at 15:02
  • $\begingroup$ I'm in highschool, haven't learns l'Hopital's rule $\endgroup$ – Ric Br Jun 19 '15 at 15:07
  • $\begingroup$ Then apply what hHhh has written below. $\endgroup$ – Simon S Jun 19 '15 at 15:07
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    $\begingroup$ I'm wondering if this is supposed to be an exercise in asymptotics. For large x, e^x dominates x and the +2 is similarly irrelevant, leaving only log_e(e^x) / x, which is 1. Not at all thorough, but good for building intuition. $\endgroup$ – MattPutnam Jun 19 '15 at 15:44
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I don't know if this is enough for you but have you realized that:

\begin{equation} \log_e(e^x + x) = \log_e(e^x(1+xe^{-x})) = x + \log_e(1 + xe^{-x}) \end{equation}

?

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L'Hospital: $$\lim_{x\to\infty}\frac{2+\ln (x+\exp x)}{x}=\lim_{x\to\infty}\frac{0+\frac{1+e^x}{x+e^x}}{1}=\lim_{x\to\infty}\frac{0+e^x}{1+e^x}=\lim_{x\to\infty}\frac{e^x}{0+e^x}=1$$

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i hope you know that $$\lim_{x\to \infty} \frac x{e^x} = 0 \text{ equivalently } \lim_{x\to \infty} \frac{e^x}{x} = \infty$$ to conclude that $$e^x >>x \text{ for large } x.$$

therefore $$\ln(e^x + x) = \ln(e^x) + \cdots = x+\cdots \text{ for large }x. $$ therefore $$ \lim_{x\to\infty}{\ln(e^x+x)+2\over x} = \lim_{x \to \infty}\frac{x+2+\cdots}{x} = 1.$$

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