Prove that $\lim_\limits{x\to 0}{\frac{e^x-1}{x}}=1$ without derivatives

We can assume the "Important limit" $\lim_{x\rightarrow 0}(1+x)^{1/x}=e$. Then \begin{align*} \lim_{x\rightarrow 0} \frac{e^x-1}{x} &= \lim_{y\rightarrow 0}\frac{y}{\ln (1+y)}\\ &= \lim_{y\rightarrow 0}\frac{1}{\ln \left((1+y)^{1/y}\right)}\\ &= 1. \end{align*}

Using $$e^x=\lim_{n\to\infty}(1+\frac{x}{n})^n=1+x+O(x^2)$$ Which is not directly using the Taylor Series, just the binomial expansion.

You can procede like in the proof of using the Taylor series.

Since $1 + x \le e^x$ for all $x$ and we have $e^x \le 1/(1 - x)$ for $x < 1$ so $$1 \le (e^x - 1)/x \le 1/(1 - x) \to 1$$ as $x\to 0$.