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what is the limit as x--> infinity of (14x^5 -81x^3-8x) / (112+31x^2-16x^4- 15x^5)? Also, what is the derivative of f(x)= (3x^5)e^x ?

2007-01-28 12:02:13 · 4 answers · asked by nyadastar 2 in Science & Mathematics Mathematics

4 answers

what is the limit as x--> infinity of (14x^5 -81x^3-8x) / (112+31x^2-16x^4- 15x^5)? Take derivative of numerator and denominator:
=(70x^4 -243x^2 -8)/(62X -64x^3 -75x^4). Do it again
= (280x^3- 486x)/(62-192x^2- 300 x^3). Do it again
=(840x^2 -486)/(384x-900 x^2) Do it again
=1680 x/(384 -1800x) Once more
= 1680/-1800 = -.9333333333
So as x --> +/ - infinity f(x) --> -.933333333333





Also, what is the derivative of f(x)= (3x^5)e^x ?

2007-01-28 12:35:11 · answer #1 · answered by ironduke8159 7 · 0 0

To find the limit on a fraction, look which has the greatest exponent. The greatest exponent is 5. So ignore everything else except 14x^5/-15x^5
Cancel out the x^5
The limit is -(14/15)

The derivative of f(x) is [(3x^5)e^x]+[(15x^4)e^x] by Product Rule.
Leave (3x^5) alone, multiply by derivative of e^x which is e^x.
Add the product of e^x times derivative of (3x^5) which is
15x^4 by exponential differentiation. (Multiply exponent by coefficient, subtract one from exponent) You can leave it as an addition or simplify.

2007-01-28 20:11:55 · answer #2 · answered by Anonymous · 0 0

-14/15. For F(x) = P(x)/Q(x), where P and Q are polynomials of the same degree,
lim x->infty F(x) = (leading coefficient of P)/(leading coefficient of Q).

Product Rule: 3(5x^4e^x+x^5e^x) = 3e^x(x^5+5x^4)

2007-01-28 20:08:57 · answer #3 · answered by lotrgreengrapes7926 2 · 0 0

my answer is at this link
http://img406.imageshack.us/img406/7228/untitledck3.jpg

2007-01-28 20:29:55 · answer #4 · answered by M. Abuhelwa 5 · 0 0

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