Answer
$$L(x)= x$$
Work Step by Step
Given $$f(x) =e^{-2x}\sin x ,\ \ a=0$$
Since $f(a) =0$ and
\begin{align*}
f'(x)&= \frac{d}{dx}\left(e^{-2x}\right)\sin \left(x\right)+\frac{d}{dx}\left(\sin \left(x\right)\right)e^{-2x}\\
&= -2e^{-2x}\sin \left(x\right)+e^{-2x}\cos \left(x\right)\\
f'(0)&= 1
\end{align*}
Then the linearization is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&=x
\end{align*}
