$\displaystyle Lim X- 0$,

$\displaystyle e^-.5x (sinx)$

I think the answer is zero, so the limit does not exist. Is this right?

Thanks

Jason

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- Jun 8th 2009, 01:01 PMDarkhrse99I need help with a limit problem!
$\displaystyle Lim X- 0$,

$\displaystyle e^-.5x (sinx)$

I think the answer is zero, so the limit does not exist. Is this right?

Thanks

Jason - Jun 8th 2009, 01:08 PMAmer
$\displaystyle lim_{x\rightarrow 0} ( e^{\frac{-x}{2}} sinx)=lim_{x\rightarrow o}\left(\frac{sinx}{e^{.5x}}\right)=\frac{0}{1}=0$