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Is there a way to find this limit without using L'Hôpital's rule. Just by using some basic limit properties.

$$\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$$

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closed as unclear what you're asking by Did, user137731, Adam Hughes, Shailesh, iadvd Sep 29 '16 at 0:24

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ How do you define $e$? $\endgroup$ – Workaholic Sep 28 '16 at 16:24
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    $\begingroup$ What tools do you have available? $\endgroup$ – Mark Viola Sep 28 '16 at 16:24
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    $\begingroup$ This is one of the ways of defining $e$. You can see the proof of the equivalence of the various definitions at ProofWiki. $\endgroup$ – user137731 Sep 28 '16 at 16:28
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    $\begingroup$ A reasonable interpretation of the question would be, how do you prove that the natural log of the given limit equals 1, where the natural log function is defined as an integral of 1/t dt? Is that what you're looking for? $\endgroup$ – G Tony Jacobs Sep 28 '16 at 16:42
  • $\begingroup$ $\lim_{x \to \infty}{x \choose k}{1 \over x^{k}} = {1 \over k!}$. $\endgroup$ – Felix Marin Sep 29 '16 at 4:38
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We may show that $$ \lim_{x\to +\infty} x\left(\log(x+1)-\log(x)\right) = 1 \tag{1}$$ by noticing that $$ x\left(\log(x+1)-\log(x)\right)=\int_{x}^{x+1}\frac{x}{t}\,dt \tag{2}$$ is trivially bounded between $\frac{x}{x+1}$ and $1$, since $f(t)=\frac{1}{t}$ is a decreasing function on $\mathbb{R}^+$.

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    $\begingroup$ Wow, this is slick. I don't think I've ever seen limits evaluated by bounding an equivalent integral.. $\endgroup$ – mike van der naald Sep 28 '16 at 16:49
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In the comments it has been pointed out that this usually serves as a definition of e. But I think I know what you're asking. Usually in introductory calculus classes to evaluate this limit you let $$y= \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x,$$ and then take the natural log of both sides to get $$\ln(y) = \lim_{x\to\infty} x\ln\left(1+\frac{1}{x}\right)=\lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}},$$ where the last limit you use L'Hôpital's rule to get that it is equal to 1. So I think maybe what you are actually asking is can, $$\lim_{x\to\infty} x\ln\left(1+\frac{1}{x}\right)$$ be evaluated without L'Hôpital's rule? Please correct me if I'm wrong. If this is indeed your question you can expand $\ln\left(1+\frac{1}{x}\right)$ as a series at $x=\infty$ which I think makes the limit trivial.

I don't think I've ever used these kinds of power series so I would appreciate any input about this last suggestion.

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  • $\begingroup$ Yes i've tried to say so. I should take a look at Taylor's series. $\endgroup$ – user373239 Sep 28 '16 at 17:31
  • $\begingroup$ Taylor series are definitely really useful. Jack's answer below is super cool and pretty elementary and shows the same limit. $\endgroup$ – mike van der naald Sep 28 '16 at 17:43
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Here are two of my questions which, combined, answer your question:

Is this proof that $\lim_{n \to \infty} (1+1/n)^n$ exists (1) new, (2) interesting?

Is showing $\lim_{z \to \infty} (1+\frac{1}{z})^z$ exists the same as $\lim_{n \to \infty} (1+1/n)^n$ exists

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