To find the limit $\lim_{{h \to 0}} \frac{e^{2h} - 1 - 2h}{h^2}$, we can use the Taylor series expansion of $e^{2h}$ around $h = 0$:

$e^{2h} \approx 1 + 2h + \frac{(2h)^2}{2!} + \frac{(2h)^3}{3!} + \cdots$ $e^{2h} \approx 1 + 2h + 2h^2 + \frac{8h^3}{6} + \cdots$ $e^{2h} \approx 1 + 2h + 2h^2 + \frac{4h^3}{3} + \cdots$

Now, substitute this approximation into the given expression:

$\frac{e^{2h} - 1 - 2h}{h^2} \approx \frac{(1 + 2h + 2h^2 + \frac{4h^3}{3} + \cdots) - 1 - 2h}{h^2}$ $\approx \frac{2h^2 + \frac{4h^3}{3} + \cdots}{h^2}$ $\approx \frac{2h^2}{h^2} + \frac{\frac{4h^3}{3}}{h^2} + \cdots$ $\approx 2 + \frac{4h}{3} + \cdots$

As $h \to 0$, all terms involving $h$ and higher powers of $h$ tend to zero, so the limit becomes:

$\lim_{{h \to 0}} \left( 2 + \frac{4h}{3} + \cdots \right) = 2$

Therefore, the limit is:

$\boxed{2}$

Do you want more details or have any questions?

Here are some related questions for further practice:

1. Evaluate $\lim_{{h \to 0}} \frac{e^h - 1 - h}{h^2}$.
2. Evaluate $\lim_{{h \to 0}} \frac{e^{3h} - 1 - 3h}{h^2}$.
3. Use the Taylor series to find $\lim_{{h \to 0}} \frac{\sin(h) - h}{h^3}$.
4. Evaluate $\lim_{{h \to 0}} \frac{\cos(h) - 1 + \frac{h^2}{2}}{h^4}$.
5. Evaluate $\lim_{{h \to 0}} \frac{\ln(1+h) - h}{h^2}$.
6. Use the Taylor series to find $\lim_{{h \to 0}} \frac{e^{4h} - 1 - 4h}{h^2}$.
7. Find $\lim_{{h \to 0}} \frac{e^{5h} - 1 - 5h}{h^2}$.
8. Evaluate $\lim_{{h \to 0}} \frac{e^{6h} - 1 - 6h}{h^3}$.

Tip: When dealing with limits involving exponential functions, using the Taylor series expansion can simplify the computation and help in identifying the leading terms that contribute to the limit.