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Evaluate $\lim_{k\to\infty} (1+\frac1k)^k$ Is there another approach to this solution other than l'Hôpital. Here's my working till now: $$\ln\left(1+\frac1k\right)^k=k\ln\left(1+\frac1k\right).$$ As $k\to\infty$, $\ln(1+\frac1k)\to0$ due to the fact that $\ln(1)=0$.

MickG
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Abmon98
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  • Related: http://math.stackexchange.com/questions/612155/is-this-true-11-nn-11-11-21-31-4-cdots-1-n – Simon S Nov 14 '15 at 20:20
  • Damn mobile keyboard, the edit summary should have been "introduced LaTeX into post; corrected spelling of L'Hôpital". – MickG Nov 14 '15 at 20:25
  • Sorry for presenting such a bad quality.I know nothing about using LaTex. – Abmon98 Nov 14 '15 at 20:55

2 Answers2

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$$\left(1+\frac1k\right)^k=e^{k\log\left(1+\frac1k\right)}$$

Using the integral definition of the logarithm function, $\log (1+z)\equiv\int_1^{1+z}\frac{1}{u}\,du$, it is easy to see that the logarithm function satisfies the inequalities

$$\frac{z}{1+z}\le \log(1+z)\le z$$

Therefore, setting $z=1/k$, we have

$$\frac{k}{k+1}\le k\log\left(1+\frac1k\right)\le 1$$

Using continuity of the exponential function along with the squeeze theorem reveals

$$\lim_{k\to \infty}\left(1+\frac1k\right)^k=e$$

Mark Viola
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  • I have not yet still been introduced to integrals. okay i know what the squeeze theorm is but why does log lie within these two boundaries. – Abmon98 Nov 14 '15 at 20:42
  • Well, help me to understand what you've studied so that we use only those tools you have at your disposal. Else, I'm guessing how you're expected to proceed. Do you know the mean value theorem and how to take the derivative of the log? – Mark Viola Nov 14 '15 at 22:50
  • What i have studied so far limits,continuity, derivatives of exponential s and logarithms,trigonometric functions and their inverse functions that's it. – Abmon98 Nov 14 '15 at 23:23
  • Do you know the derivative of the log function? – Mark Viola Nov 14 '15 at 23:26
  • yes the derivative of ln(x)=1/x – Abmon98 Nov 14 '15 at 23:30
  • OK. Then, what is the limit $\lim_{k\to \infty}k\log \left(1+\frac1k\right)$ if you change variables and let $k=1/h$? – Mark Viola Nov 14 '15 at 23:36
  • lim 1/h-->0 1/hlog(1+h)=lim1/h-->0log(1+h)/h thats the slope of the log at x=1, not sure though. – Abmon98 Nov 14 '15 at 23:41
  • Yes. And you know the slope is given by the ... – Mark Viola Nov 14 '15 at 23:43
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As $h\to 0,$ $[\ln (1+h)]/h = [\ln (1+h) - \ln 1]/(h-0) \to \ln '(1) = 1.$ This is just from the definition of the derivative; no L'Hopital was used.

zhw.
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  • I get that this comes from the definition of a derivative but why is ln(1+h)/h-->1 as h-->0. – Abmon98 Nov 14 '15 at 22:16
  • I assumed we know $\ln'(x) = 1/x.$ – zhw. Nov 14 '15 at 22:55
  • @zhw. I'm not sure as to which tools the OP has. Often, the limit is taken as the definition of $e$. And it is easy to show that $\left(1+\frac1k \right)^k$ is increasing monotonically and bounded above by $3$ (and below by $2$. – Mark Viola Nov 14 '15 at 23:30