No L'Hôpital or tables allowed here!

$\lim_{x\to 0} \frac{e^x-1}{x}=\lim_{x\to 0} \frac{e^{i(-ix)} -1}{x}$ $=\lim_{x\to 0} \frac{\cos(ix)-i\sin(ix)-1}{x}=\lim_{x\to 0}\frac{\cos(ix)-1}{x}-i\lim_{x\to 0}\frac{\sin(ix)}{x}$ $=\lim_{x\to 0}\sin(ix)\cdot \frac{\sin(ix)}{x(\cos(ix)+1)}-i^2\lim_{x\to 0}\frac{\sin(ix)}{ix}=i\lim_{x\to 0}\sin(ix)\cdot \frac{\sin(ix)}{ix(\cos(ix)+1)}+1\cdot 1$ $i\cdot 0 +1=1$

let f(x)= $e^x$ , so

$\lim_{x \rightarrow0}\frac{e^x-1}{x}=\lim_{x \rightarrow0}\frac{f(x)-f(0)}{x-0}$

Observe that the limit above is the definition of derivative of f(x) at x=0, f'(x)= $e^x$ f'(0)=1

so, $\lim_{x \rightarrow0}\frac{e^x-1}{x}$ =1

By using Taylor series expansion of e^x.

Well A common problem for JEE students And almost everyone does by Taylor series expansion ..

Use the expansion of e^x by taylor