lim as x - 0+ for x/(ln(1-x^3))

Question

lim as x -> 0+ for x/(ln(1-x^3))

anonymous · Answer

Do I use L'Hopital or should I do it normally?

anonymous · Answer

l'Hopital's since it it 0/0

anonymous · Answer

That gives me 0, agreed? Or am I missing something?

anonymous · Answer

0/0 does not equal zero.

nor does the limit as x->0+ after applying lHopital's

anonymous · Answer

$$\frac{ f'(x) }{ g'(x) } = \frac{x^3 - 1 }{ 3x^2 }$$

anonymous · Answer

What do I do when I get to -1/0 then? I guess that's where I got confused and for some reason just wrote it as equal to 0..

anonymous · Answer

-1/0+ = - infinity = -1/0-

you can check on your calculator that 1/±0.0000000001 = a big number.

as the denominator gets infinitely smaller, 1/it approaches infinity

anonymous · Answer

Ahhhhh. Apply it vice subbing in 0 all the time, aka think not plug-n-chug. Thank you

anonymous · Answer

glad i could help :)