$\displaystyle Lim X- 0$, $\displaystyle e^-.5x (sinx)$ I think the answer is zero, so the limit does not exist. Is this right? Thanks Jason
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$\displaystyle lim_{x\rightarrow 0} ( e^{\frac{-x}{2}} sinx)=lim_{x\rightarrow o}\left(\frac{sinx}{e^{.5x}}\right)=\frac{0}{1}=0$
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