Find f'(0) by definition for f(x) = e^x

Find $f'(0)$ by definition for $f(x) = e^{x}$.

Step 1: Write the definition of the derivative at $x = 0$

\[ f'(0) = \lim_{\Delta x \to 0} \frac{f(0 + \Delta x) - f(0)}{\Delta x}. \]

Step 2: Compute the increment $\Delta y$

\[ f(0 + \Delta x) = e^{\Delta x}, \qquad f(0) = e^{0} = 1, \]

so that

\[ \Delta y = e^{\Delta x} - 1. \]

Step 3: Form the difference quotient

\[ \frac{\Delta y}{\Delta x} = \frac{e^{\Delta x} - 1}{\Delta x}. \]

Step 4: Evaluate the limit

We use the standard limit

\[ \lim_{u \to 0} \frac{e^{u} - 1}{u} = 1. \]

Setting $u = \Delta x$ gives

\[ f'(0) = \lim_{\Delta x \to 0} \frac{e^{\Delta x} - 1}{\Delta x} = 1. \]

Step 5: Verification

Using the standard rule $\bigl(e^{x}\bigr)' = e^{x}$: \[f'(x) = e^{x} \;\Longrightarrow\; f'(0) = e^{0} = 1 \;\;\checkmark\]

The same conclusion follows from the series $e^{u} = 1 + u + \dfrac{u^{2}}{2!} + \cdots$, since

\[ \frac{e^{u}-1}{u} = 1 + \frac{u}{2!} + \frac{u^{2}}{3!} + \cdots \;\xrightarrow[u \to 0]{}\; 1. \]