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OpenStudy (anonymous):
lim x->0 (2cosx+3x-2)/5x what do you guys think? I think it is 2/5, but someone says it is 3/5
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OpenStudy (anonymous):
If you substitute in x = 0 you get (2cos(0)+3(0)-2)/5(0). This simplifies to (2-2)/0 or 0/0 one of the L'Hopital equations. Taking the derivative yields (-2sin(x)+3)/5. Now putting in 0 gets 3/5. So I agree with the 3/5.
OpenStudy (anonymous):
Oh I see thank you my friend. I had a calculation error
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