Question

LIMITS

Answer 1

Limits are pretty cool, yeah.

Answer 2

What you want to know about limits?

Answer 3

\[\lim_{x \rightarrow 0} (x^2 + 2\cos x - 2)/(x \times \sin^3 x)\]
how to evaluate this???

Answer 4

\[\frac{x^2+2\cos x-2}{x\sin^3x}=\frac{x%^2}{x\sin^3x}+\frac{2\cos x}{x\sin^3x}-\frac{2}{x\sin^3x}\]Maybe this helps?

Answer 5

Also, remember the liberal application of l'Hopital's rule:\[\left\{\lim\frac{f\left(x\right)}{g\left(x\right)}\mid\lim f\left(x\right)=\lim g\left(x\right)=0,\pm\infty\right\}\iff\lim\frac{f\left(x\right)}{g\left(x\right)}=\lim\frac{f'\left(x\right)}{g'\left(x\right)}\]

Answer 6

if i apply L-hospital the denominator repeatedly comes as zero, just see??

Answer 7

\[\frac{2}{x\sin^3x}=\frac{2x^{-1}}{\sin^3 x}\]Remember this as well.

Answer 8

Also, \[\lim f\cdot g=\lim f\cdot\lim g\]

Answer 9

Yeah, I think you can solve it with this info.

Answer 10

power series method works nice.

Answer 11

whats that???

Answer 12

Have you done the taylor series?

Answer 13

no!!!

Answer 14

ok..then ignore what I typed above

Answer 15

but please help???

Answer 16

just use l'hospitals rule then

Answer 17

you will have to do it several times

Answer 18

using l hospitals rule,twice,the denominator becomes very nasty!!!
please tell me about the taylor series...

Answer 19

http://en.wikipedia.org/wiki/Taylor_series

Answer 20

how do you apply it here???