Derivatives by First Principles

L'hopital's Rule.

$\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{\frac{d}{dx} (e^x - 1)}{\frac{dx}{dx}} = \lim_{x \to 0} e^x = e^0 = \boxed{1}$

$\begin{aligned} \lim_{x \to 0} \frac{\color{#3D99F6}{e^x}-1}{x} & = \lim_{x \to 0} \frac{\color{#3D99F6}{\left(1+x+\frac{x^2}{2}+ \frac{x^3}{6}+ ...\right)}-1}{x}\quad \quad \small \color{#3D99F6}{\text{Using Maclaurin series}} \\ & = \lim_{x \to 0} \space 1+\frac{x}{2}+ \frac{x^2}{6}+ ... \\ & = \boxed{1} \end{aligned}$

The derivative of $f(t)=e^t$ at $t=0$ is $f'(0)=\lim_{x\to 0}\frac{e^x-e^0}{x-0}=\lim_{x\to 0}\frac{e^x-1}{x}=e^0=\boxed{1}$

Del' Hospital rules this one too

I did it without L'Hopital's. $e^{x} = \sinh x + \cosh x$

$= \displaystyle \lim_{x \rightarrow 0} \frac{\sinh x + \cosh x - 1}{x}$ $= \displaystyle \lim_{x \rightarrow 0} \frac{(\sinh x + \cosh x - 1)(\sinh x - \cosh x -1)}{x(\sinh x - \cosh x - 1)}$ $= \displaystyle \lim_{x \rightarrow 0} \frac{\sinh^{2}x + 1 -2\sinh x - \cosh^{2} x}{x(\sinh x - \cosh x -1)}$

Using pythagorean identity for hyperbolic trig functions, we know that $\sinh^{2} x - \cosh^{2} x = -1$ .

$= \displaystyle \lim_{x \rightarrow 0} \frac{ -2\sinh x }{x(\sinh x - \cosh x -1)}$

We know that $\displaystyle \lim_{x \rightarrow 0 } \frac{\sinh x}{x} =1$ .

$= \displaystyle \lim_{x \rightarrow 0} \frac{-2}{\sinh x - \cosh x - 1}$

$\sinh(0) = 0, \cosh(0) = 1; \lim_{x \rightarrow 0} \frac{-2}{-2} = \boxed{1}$

This is the method I used(simple problem)

L'Hopital's rule.