Find the bound of the absolute error of the approximation
Find the bound of the absolute error of the approximation $e^x\approx1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^{n-1}}{(n-1)!}$, $0\lt x\lt 1$.
Explanation of Method
Bounding e to the first power on an open unit interval hands the whole error to a single factorial.
Step 1: Identify the remainder
The given polynomial is the Maclaurin polynomial of degree $n-1$ for $e^x$; with $f^{(n)}(x)=e^x$, \[e^x = \sum_{k=0}^{n-1}\frac{x^k}{k!}+R_{n-1}(x), \qquad R_{n-1}(x)=\frac{e^\xi}{n!}x^n, \tag{1}\] for some $\xi$ strictly between $0$ and $x$.
Step 2: Bound the remainder on $(0,1)$
Since $0\lt \xi\lt x\lt 1$: $e^\xi\lt e^x\lt e^1=e\lt 3$; also $x^n\lt 1$. Hence
\[ |R_{n-1}(x)| = \frac{e^\xi}{n!}x^n \lt \frac{e}{n!} \lt \frac3{n!}. \tag{2} \]
Step 3: Verification
\[n=5,\ x=0.99\text{: actual error}\approx0.009443,\ \text{bound }e/5!\approx0.022652\ \checkmark\ \text{(error stays under the bound throughout }(0,1)\text{)}\]
Result
Final answer:
\[\boxed{\,\left|e^x - \left(1+x+\frac{x^2}{2!}+\cdots+\frac{x^{n-1}}{(n-1)!}\right)\right| < \frac{e}{n!} < \frac3{n!}, \qquad 0<x<1\,}\]
This bound is uniform over $(0,1)$ (approached, never attained, as $x\to1^-$):
Since $e/n!\to0$ as $n\to\infty$, the Maclaurin series converges to $e^x$ uniformly on $(0,1)$.