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The limit $\lim\limits_{x\to0}(1+x)^{\frac{1}{x}}=e$ can be proved using La Hôpital's Rule but is it possible to prove this limit using the definition of the limit?

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    @Surb Well, it was defined, in my book as $$e=\lim_{x \to 0} (1+x)^{\frac{1}{x}}=\lim_{t \to \infty} \left(1+\frac{1}{t}\right)^t$$ – 2017-02-22
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    @S.C.B.: If in one of your book is defined as the first limit, it's not an usual definition. But indeed, the second is the definition of $e$. – 2017-02-22
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    @Surb OK, that was a typo. – 2017-02-22
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    see here http://aleph0.clarku.edu/~djoyce/ma122/elimit.pdf i hope this will help you – 2017-02-22
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    What is your definition of $e$? If it is $\lim_{x\to +\infty}\left(1+\frac{1}{x}\right)^x$ there is nothing to prove, if it is $\lim_{n\to +\infty}\left(1+\frac{1}{n}\right)^n$ there is little to prove, if it is $\sum_{n\geq 0}\frac{1}{n!}$ there actually are some things to prove. – 2017-02-22

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Set $y=\frac{1}{x}$, and thus $$\lim_{x\to 0^+}(1+x)^{1/x}=\lim_{y\to +\infty }\left(1+\frac{1}{y}\right)^y.$$

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    But then how will you prove $\lim \limits_{y\to +\infty }\left(1+\frac{1}{y}\right)^y$? – 2017-02-22
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    @Razin: This is the definition of $e$ ! [See here](https://en.wikipedia.org/wiki/E_(mathematical_constant)) – 2017-02-22
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    @Surb: truth to be told, $e$ is usually defined as the limit of a sequence, not the the limit of a function. Still, there is little to prove. – 2017-02-22
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Assume the limit exists and the value is some $l\in\mathbb{R}$, that is $$\lim\limits_{x\to0}(1+x)^{\frac{1}{x}}=l.$$ Consider the logarithm in both sides of the expression, $$\log\left(\lim\limits_{x\to0}(1+x)^{\frac{1}{x}}\right)=\log l.$$ This is equivalent to $$\lim\limits_{x\to0}{\frac{1}{x}}\log(1+x)=\log l.$$ Now, since $\log(1) = 0$, we can write ${\frac{1}{x}}\log(1+x)$ as $${\frac{1}{x}}\log(1+x) = {\frac{\log(1+x)-\log(1)}{x}},$$ and then, by the definition of the derivative, the limit on the left hand side is equal to $(log(x))'|_{x=1} = 1$. Thus, $$1=\log l,$$ which means that the value $l$ is $e$.

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    I think you should clarify that the limit on the left hand side is equal to $1$ by the very definition of derivative, since one might think that l'Hopital is being used. – 2017-02-22
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    @AloizioMacedo You are right. Explanation added. – 2017-02-22
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Consider $$A=(1+x)^{\frac{1}{x}}\implies \log(A)=\frac{1}{x}\log(1+x)$$ Now, remembering equivalents for small $x$ $$\log(1+x)\sim x$$ or Taylor series $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ you arrive to $$\log(A)\sim 1$$ or $$\log(A) =1-\frac{x}{2}+O\left(x^2\right)$$

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If you don't define $e$ as that limit, you can do what follows:

$$\left( 1 + \frac1{x}\right)^x = \exp\left( x \ln \left(1 + \frac1{x}\right) \right) = \exp\left( x \cdot \left( \frac1{x} + O\left(\frac1{x} \right) \right) \right) = e \cdot \exp( O(1)) \to e$$