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Let $a\gt0$ real. Evaluate the following limit: $$\displaystyle\lim_{n\to\infty}\left(a+\tfrac1n\right)^n $$ I try'd expressing this limit in such a way that I could use the known limit: $\displaystyle\lim_{n\to\infty}\left(1+\tfrac1n\right)^n = e $. However that hasn't really helped me solve the problem. Does anyone have another idea?

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    $\begingroup$ Please refrain from using displaystyle MathJax in the question title. the appropriate MathJax is $\text{\limits}$ $\endgroup$ – AlexR Jan 12 '14 at 21:09
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$$\begin{align*}(1)&\;\;a>1\implies\left(a+\frac1n\right)^n=a^n\left(1+\frac1{an}\right)^n\stackrel{\text{Bernoulli's Ineq.}}\ge a^n\left(1+\frac1a\right)\xrightarrow[n\to\infty]{}\infty\\(2)&\;\;a<1\implies\left(a+\frac1n\right)^n=a^n\left(1+\frac1{an}\right)^n\xrightarrow[n\to\infty]{}0\end{align*}$$

since

$$\left(1+\frac{1/a}{n}\right)^n\xrightarrow[n\to\infty]{}e^{1/a}$$

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    $\begingroup$ Bernoulli's inequality is overkill: all you need is $(1+\frac{1}{an})^n > 1$. $\endgroup$ – TonyK Jan 12 '14 at 22:07
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    $\begingroup$ "Overkill"? Bernoulli's inequality is sometimes studied in high school, so I don't think I'd consider it overkill, and yes: it's enough to have the inequality you mention. $\endgroup$ – DonAntonio Jan 12 '14 at 22:13
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If $a > 1$, then $$\left(a + \frac 1 n\right)^n > a^n$$ which blows up as $n$ grows.

If $a < 1$, then $a + \frac 1 n$ can be made less than $1 - \epsilon$ for some very small error $\epsilon$ and large $n$; what happens when you raise this to a large power?

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  • $\begingroup$ Well if it can be made less than 1, it will go to zero if raised to a large number. Thanks for answering, this made it pretty clear. $\endgroup$ – Spacer Jan 12 '14 at 21:16

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