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lim(x->0) (sin8x)/x
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l'hopital's rule...
yes
I think it is 8
have to find the limit
thank you
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\[\frac{\frac{d(\sin8x)}{dx}}{\frac{dx}{dx}}=8\cos(8x)=8\]
That's the lim as x approaches 0 of course. I just don't know how to write that.
you can also do this myininay way
Let's see it.
\[8\times \frac{\sin(8x)}{8x}\] \[8\times 1\]
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What else does that apply to?
not sure what you mean exactly, but you could use this for \[\lim_{x\to 0}\frac{\sin(ax)}{bx}\]
given of course that we grand that \[\lim_{x\to 0}\frac{\sin(ax)}{ax}=1\] adn could also use it for \[\lim_{x\to 0}\frac{\sin(ax)}{\sin(bx)}\]
You guys are way too smart for me. I'm just an old Michigan farm girl.
why am i not convinced?
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I can't imagine.
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