limits in indeterminate forms?

Question

how would you evaluate lim (3x^7 + 4x^2) / (9x^7 - x^3 + 2)
x=>oo

btw, thats as x approaches infinity.
'Thanks!

i originally thought it was 1/3 too, but that rule doesn't work if the limit is indeterminate

dont know why, but my calc teacher says so

Answer 1

Why not? If you divide top and bottom of this expression by x^7 you get
(3 + 4/(x^5))/(9 - 1/(x^4) + 2/(x^7))

Every term except the 3 and the 9 now has a power of x as denominator, and therefore approaches 0 as x goes to infinity,
therefore it approaches 3/9.

Answer 2

lim (3 + 4x^-5)/(9 - x^-4 + 2^-7) = 3/9 = 1/3.
x→∞

Looks to me that we've arrived at the same answer with 2 diffeerent methods. I would be interested in knowing by what reasoning your calc teacher says both are invalid.

Answer 3

Method I : Take x^7 common from numerator and denominator and then put the value of x as infinity the answer will come out as 1/3.

Method II : U can apply L-Hospital rule

Answer 4

(3x^7 + 4x^2) / (9x^7 - x^3 + 2)
= [(9x^7 - x^3 + 2)/3 + ((x^3)/3 + 4x^2 - 2/3)] / (9x^7 - x^3 + 2)
= 1/3 + ((x^3)/3 + 4x^2 - 2/3) / (9x^7 - x^3 + 2)

As x approaches infinity, the 2nd term approaches zero, so the expression approaches 1/3.

I bet you misunderstood your teacher :)

Answer 5

You can apply l'Hospital's rule, and differentiate both numerator and denominator successively until it becomes determinate. You will be left with 3/9 = 1/3.

Answer 6

1/3
use l'hôpital's rule three times.
n - numerator ; d - denominator

n' = 21x^6 + 8x ; d'=63x^6 - 3x^2
n'' = 126x^5 + 8 ; d'' = 378x^5 - 6x
n''' = 630x^4 ; d''' = 1890x^4

n'''/d''' = 630x^4 / 1890x^4 = 1/3

Answer 7

1/3

because the other terms become negligible as x becomes large.
Only the x^7 terms are relevant