The image shows a limit problem: $\underset{x\to 0}{\lim }\frac{e^{-2x}-1}{x}$. This is a standard indeterminate form 0/0. To solve it, we recognize it as the definition of the derivative of $e^{-2x}$ at x=0. The derivative of $e^u$ is $e^u\frac{du}{dx}$. Here, u = -2x, so the derivative is $e^{-2x}\times (-2)$. Evaluating this at x=0 gives $e^0\times (-2)=1\times (-2)=-2$.

Final Answer: $-2$