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Precalculus Mathematics Homework Help
Binomial theorem to evaluate limits?
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[QUOTE="HallsofIvy, post: 4534153, member: 637751"] You are fine to lim h->0 ((h+1)^9 + h-1)/((h+1)^4+h-1) But I don't understand why you then switched to "1/h". Taking h to 0 is much simpler than taking 1/h to infinity. By the binomial theorem, (h+ 1)^9= h^9+ 9h^8+ ...+ 36h^2+ 9h+ 1 so that (h+1)^9+ h- 1= h^9+ 9h^8+ ...+ 36h^2+ 10h. Similarly (h+ 1)^4= h^4+ 4h^3+ 6h^2+ 4h+ 1 so that (h+ 1)^4+ h- 1= h^4+ 4h^3+ 6h^2+ 5h. The fraction can be written [tex]\frac{h^9+ 9h^9+ ...+ 36h^2+ 10h}{h^4+ 4h^3+ 6h^2+5h}= \frac{h(h^8+ 9h^7+ ...+ 36h+ 10)}{h(h^3+ 4h^2+ 6h+ 5)}[/tex] Now those first "h" terms cancel and you can just take h= 0. Are you [b]required[/b] to use that method? Simpler than that or L'Hopital: The only problem here is that both numerator and denominator are 0 at x= 1. But since they are polynomials, that means that each has x- 1 as a factor: [tex]x^9+ x- 2= (x- 1)(x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2)[/tex] and [tex]x^4+ x- 2= (x- 1)(x^3+ x^2+ x+ 2)[/tex] So you just need to evaluate [tex]\frac{x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2}{x^3+ x^2+ x+ 2}[/tex] at x= 1. [/QUOTE]
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Binomial theorem to evaluate limits?
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