Verified answer

Since lim x→0 (1-e^3x)/tanx has a 0/0 form, we can use L'Hopital's Rule:

lim x→0 (1-e^3x)/tanx

= lim x→0 derivative of (1-e^3x) / derivative of tanx

= lim x→0 (-3e^3x) / sec^2 x, from using the chain rule for the numerator

= (-3e^(3*0)) / sec^2 (0)

= (-3*1)/(1^2)

= -3.

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