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chapset said:
dany612 said:
pearljammer said:
ShadowSnake said:

Secondly to some of the responses being fed your way. If I remember correctly, you should not be telling dany about l'hopitals rule as that is a long ways from being taught to him.

Ah, you're right! I just checked and I hadn't been taught L'Hopital's rule until second semester of my first year.

so theres a more simplier way to solve it without the l'hospital rule?

L'Hopital is the simplest way, you do a derivative on top and a derivative in the bottom of your fraction, then you evaluate your limit

ex: lim x->0   (x^2 + 4x)/(x +4)

derivative on top = 2x + 4

derivative in the bottom = 1

then you evaluate your limit

lim x->0 (2x + 4)/1 = [2(0)+4]/1 = 4

 my english is not so great I hope you understand a bit better

Actually, I don't think L'Hopital's rule applies here. IIRC, it's a rule as a last resort; only use it if you can't simplify it beyond its current form.

 

lim x->0 (x^2 + 4x)/(x + 4)  can be simplified:

lim x->0 [x(x+4)]/(x+4)

lim x->0 (x) = 0



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