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I know that in the very small values of $x$ $$(1+x)^n ≈1+ nx$$ and I can prove it using Taylor series.
But I wanted to prove it without any smell of derivative.

So... In order to calculate the following limit

$$\lim _{x \rightarrow\ 0} \frac{(1+x)^n-1} x, $$

I know that the result must be equal to $n$.

But is there any method without using l'Hospital's rule?

Note : n can be any real value not just for integers, so I didn't want to use binomial theorem.

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    $\begingroup$ "without any smell of derivative" - I guess this means you don't want to use the definition of the derivative? That is, if $f(x)=(1+x)^n$, then$$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{(1+x)^n-1}x=n(1+0)^{n-1}=n$$ $\endgroup$ – user170231 Feb 15 at 17:17
  • $\begingroup$ It's not necessarily incorrect to use $n$ here to represent a real number, but it is impolite. $n$ without any immediate context will be assumed to denote a natural number (hence the answers below). For a fixed arbitrary real number, especially in an exponent, $\alpha$ is the symbol I've seen most authors use. $\endgroup$ – Robert Wolfe Feb 15 at 20:20
  • $\begingroup$ @user170231 I think this question could be interpreted as proving that the derivative of $z^n$ at $z=1$ is indeed $n$. In that case, that would be circular reasoning. $\endgroup$ – JiK Feb 15 at 20:45
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    $\begingroup$ @JiK: So if $f(x) = x^\alpha$ and we prove $f'(x) = {\alpha}x^{\alpha-1}$, then we're good? $\endgroup$ – jxh Feb 15 at 23:52
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Hint

$$\dfrac{(1+x)^n-1}{x}=\dfrac{1+nx+\dots+nx^{n-1}+x^n-1}{x}=n+\binom{n}{2}x+\dots+nx^{n-2}+x^{n-1}.$$

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Hint:

Let $u=1+x$. Then $$\lim_{x\to 0}\dfrac{(1+x)^n-1}{x}=\lim_{u\to 1}\dfrac{u^n-1}{u-1}. $$ Now expand $u^n-1$ and apply the limits.

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Hint: $z^n - 1 = (z-1)(z^{n-1} + z^{n-2} + \cdots + 1)$. There are $n$ terms in the second factor, so it equals $n$ when $z = 1$.

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  • $\begingroup$ I already tried that, but what if n=1/3 for example? $\endgroup$ – El Mouden Feb 15 at 17:28
  • $\begingroup$ Hmm, arbitrary $n$. You could make an argument based on $e^x \approx 1 + x$, from which it follows both $e^{nx} \approx 1 + nx$ and $e^{nx} \approx (1 + x)^n$. $\endgroup$ – Jim Ferry Feb 15 at 17:40
  • $\begingroup$ No, you just did the same thing using Taylor series of $e^x$ $\endgroup$ – El Mouden Feb 15 at 17:42
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    $\begingroup$ If you want to stay elementary, you could note that once you've established the result for integers you can say $1+x \approx (1+nx)^{1/n}$, so $1+y/n \approx (1+y)^{1/n}$, which establishes it for reciprocals of integers. You can then generalize to all positive rational numbers, and then to all rationals. Generalizing to all real numbers raises the question of what real numbers are, which gets into inherently calculus-related concepts. $\endgroup$ – Jim Ferry Feb 15 at 18:10
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Well (take it jokingly), since you want to go back the time past Taylor and Newton and Leibniz, please allow a stop in the Renaissance .
At that time $(1+x)^r$ was much in vogue as the formula of compound interest and it was of course well known that at low interest it could be simplified as $1+rx$, also for $r$ corresponding to a fractional period of time.
Then you might have had the chance to make acquaintance with J. Napier and H. Briggs and join them in discussing the best base to give to the logarithm, and choose that for which $ln(1+x) \approx x$.
But then you should have waited for some decades to learn from Jacob Bernoulli that such an optimal base was the limit of $(1+1/n)^n$.

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Let $\alpha$ be a real number strictly between $0$ and $1$. Bernoulli's Inequality asserts that $$x^\alpha\leq 1+\alpha (x-1)$$ for each nonnegative real $x$.

There are a multitude of ways that one can establish this inequality. If you are adamant about avoiding even derivatives, you can accomplish this with continuity and the case where $\alpha$ is rational (which may be proven by the Arithmetic-Geometric mean which can be proven by...). If you want to avoid even continuity, you'd use least-upper-bounds and greatest-lower-bounds.

But let me assume that you have establish Bernoulli's Inequality at least. Then it's quite easy to establish from Bernoulli's that $$\alpha(1-x^{-1})\leq x^\alpha-1\leq \alpha(x-1)$$ for any nonnegative real number $x$ (so long as $\alpha$ is strictly between $0$ and $1$). The right-hand inequality is precisely Bernoulli's. The left-hand is arrived at by using $x^{-1}$ in Bernoulli's and then using the inequality $x+x^{-1}\geq 2$.

Of course, this inequality will prove your assertion for $\alpha$ strictly between $0$ and $1$. What do you when this is not the case? Choose $k$ in $\mathbb{Z}$ such that $0<\alpha/k<1$. Then use the prior case with $\alpha/k$ to conclude $$\frac{\alpha}{k}(1-(x^k)^{-1})\leq (x^k)^{\alpha/k}-1\leq\frac{\alpha}{k}((x^k)-1)$$ which, simplified, looks like $$\frac{\alpha}{k}(1-x^{-k})\leq x^\alpha-1\leq\frac{\alpha}{k}(x^k-1)\,.$$ The integer case will finish the proof from here.

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If you take $e := \lim_{n\to\infty} (1+1/n)^n$, then one can show that $\lim_{n\to\infty} (1+y/n)^n = e^y$. In fact, one can show that $$ 1 + x \le e^x \quad\text{and}\quad e^y \le 1 + y + \epsilon(y), $$ where $\epsilon(y)$ is a term such that $\epsilon(y)/y \to 0$ as $y \to 0$. Importantly, this can all be done without the definition of a derivative; in particular, it is the type of question that can be given to a first year undergraduate in their first analysis course. Given this, one can upper bound $(1+x)$ by $e^x$ and then $e^{nx}$ by $1 + nx + \epsilon(nx)$. We then find that the limit is $n$. (Note that there $n$ is a fixed positive real---not necessarily an integer---and we take $x \to 0$; there is no "$n \to \infty$ as $x \to 0$".)

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