Question

help! medal

Answer 1

Heya heya! Make sure to post your question so we can give you the best help we can :)

Answer 2

Ah sorry I can't help you :(

Answer 3

@inkyvoyd

Answer 4

This is a L'Hospital's Rule problem (Sometimes spelled "L'Hôpital's Rule." Try to evaluate the limit directly, by substituting x = 0. That fails:
\[\lim _{x \rightarrow 0}\frac{e^{3x}-3x-1}{x^2}=\frac{0}{0}\] which is an "inderminate form." You can replace the numerator and denominator with their derivatives, then try to find the limit again.
\[\lim_{x \rightarrow 0}\frac{3e^{3x}-3}{2x} = \frac{0}{0}\] which is still an indeterminate form. Apply L'Hospital's Rule again. Replace the numerator and denominator with their derivatives.
\[\lim_{x \rightarrow 0} \frac{9e^{3x}}{2}\] Evaluate this at x = 0, and you'll get a specific number for the limit.
More info and examples: http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

Answer 5

Well, that was nice of the editor to change my post! Also spelled L'H(letter o with a circumflex over it)pital's Rule.

Answer 6

9e/2?

Answer 7

what do i replace for e? @ivycoveredwalls

Answer 8

Careful...if x = 0, then \[e^{3x}=e^{0}\] and that equals.......what?

Answer 9

0

Answer 10

so the answer would just be 0? because 9*0/2

Answer 11

Not quite! \[e^{0}=1\]

Answer 12

YAY!

Answer 13

thank you !! (:

Answer 14

You're quite welcome1