Thread: I 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

$\displaystyle lim_{x\rightarrow 0} ( e^{\frac{-x}{2}} sinx)=lim_{x\rightarrow o}\left(\frac{sinx}{e^{.5x}}\right)=\frac{0}{1}=0$