anonymous
  • anonymous
LIMITS
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Limits are pretty cool, yeah.
anonymous
  • anonymous
What you want to know about limits?
anonymous
  • anonymous
\[\lim_{x \rightarrow 0} (x^2 + 2\cos x - 2)/(x \times \sin^3 x)\] how to evaluate this???

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anonymous
  • anonymous
\[\frac{x^2+2\cos x-2}{x\sin^3x}=\frac{x%^2}{x\sin^3x}+\frac{2\cos x}{x\sin^3x}-\frac{2}{x\sin^3x}\]Maybe this helps?
anonymous
  • anonymous
Also, remember the liberal application of l'Hopital's rule:\[\left\{\lim\frac{f\left(x\right)}{g\left(x\right)}\mid\lim f\left(x\right)=\lim g\left(x\right)=0,\pm\infty\right\}\iff\lim\frac{f\left(x\right)}{g\left(x\right)}=\lim\frac{f'\left(x\right)}{g'\left(x\right)}\]
anonymous
  • anonymous
if i apply L-hospital the denominator repeatedly comes as zero, just see??
anonymous
  • anonymous
\[\frac{2}{x\sin^3x}=\frac{2x^{-1}}{\sin^3 x}\]Remember this as well.
anonymous
  • anonymous
Also, \[\lim f\cdot g=\lim f\cdot\lim g\]
anonymous
  • anonymous
Yeah, I think you can solve it with this info.
Zarkon
  • Zarkon
power series method works nice.
anonymous
  • anonymous
whats that???
Zarkon
  • Zarkon
Have you done the taylor series?
anonymous
  • anonymous
no!!!
Zarkon
  • Zarkon
ok..then ignore what I typed above
anonymous
  • anonymous
but please help???
Zarkon
  • Zarkon
just use l'hospitals rule then
Zarkon
  • Zarkon
you will have to do it several times
anonymous
  • anonymous
using l hospitals rule,twice,the denominator becomes very nasty!!! please tell me about the taylor series...
Zarkon
  • Zarkon
http://en.wikipedia.org/wiki/Taylor_series
anonymous
  • anonymous
how do you apply it here???

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