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In order to prove that the relation $\lim_{n\rightarrow\infty}(1+\frac{z}{n})^n=\exp(z)$ holds for all values of z, I first prove it for positive z by showing the lower and upper bound as taylor series for $exp(z)$. To extend the proof of this relation to hold for negative z, I need to prove $\lim_{n\rightarrow\infty} (1-\frac{z^2}{n^2})^n =1$, I use the binomial expansion of the series and write it down $\lim_{n\rightarrow\infty} \sum_{k=0}^n \frac{z^{2k}}{n^{2k}}\frac{{n}{!}}{{n-k}{!}{k}{!}}$ and further expand the combinatorial term. How do I prove that the limit of this is 1 ?

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Here is an approach.

$$ e^{{n}\ln(1-z^2/n^2)} = e^{{n}(-z^2/n^2+O(z^4/n^4))} \longrightarrow_{n\to \infty} 1 . $$

We used the Taylor series

$$\ln(1-t) = -t+O(t^2) .$$

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A much simpler technique relies on the inequality $$(1 + x)^{n} \geq 1 + nx$$ for $x > -1$ and $n$ a positive integer. You should be able to see that this is trivial to show when $x \geq 0$ using the binomial theorem for positive integral exponents.

For the case when $-1 < x < 0$ we need a bit more patience. Let $f(x) = (1 + x)^{n} - 1 - nx$ then we know that $f(0) = 0$ and $f'(x) = n(1 + x)^{n - 1} - n$. If $-1 < x < 0$ and $n \geq 1$ then we can see that $f'(x) \leq 0$ so that $f(x)$ is decreasing in $(-1, 0]$. It follows that $f(x) \geq f(0)$ for all $x \in (-1, 0]$. Hence $(1 + x)^{n} \geq 1 + nx$.

We can easily notice that as $n \to \infty$, after a certain point we will have $0 < z^{2}/n^{2} < 1$ so that if $y = -z^{2}/n^{2}$ then $-1 < y < 0$. And hence $$1 + ny \leq (1 + y)^{n} \leq 1$$ or $$1 - \frac{z^{2}}{n} \leq \left(1 - \frac{z^{2}}{n^{2}}\right)^{n} \leq 1$$ Using sqeeze theorem we now get $$\lim_{n \to \infty}\left(1 - \frac{z^{2}}{n^{2}}\right)^{n} = 1$$ The advantage of this proof is that it does not rely on any properties of exponential and logarithmic functions.

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