What is lim(e^(4x)-1 -4x)/x^2 when x approaches infinity?

Question

Pls show the complete work!!! Thanks a lot.

Answer 1

Ord(e^(4x)-1 -4x) ~ Ord(e^(4x) > Ord(x^2)
Then:

lim(e^(4x)-1 -4x)/x^2 {x--> oo} = oo

Remember

ln x < x^a < e^x < x^x

Answer 2

Work? The limit doesn't exist (is infinity), the numerator is exponential and the denominator is polynomial.

Replace x by 4y to simplify, this is
lim(e^y - 1 - y)/(16y^2) = 1/16(e^y - y - 1)/y^2
= 1/16(lim(e^y/y^2) - lim(y/y^2) - lim(1/y^2))
= 1/16(lim(e^y/y^2) - 0 - 0)
= 1/16(lim(e^y/y^2))

So if you know the exponential is bigger you are done.

If not, you can use L'Hospital:
lim(e^y/y^2) = lime^y/2y = lime^y/2 = oo

Answer 3

(e^(4x)-1 -4x)/x^2
e^4x/x^2-1/x^2-4x/x^2
e^4x/x^2-1/x^2-4/x lim x--->infinity, the 2nd & 3rd terms aproach 1/infinity & 4/infinity. Both become 0
the 1st term
e^4x/x^2 becomes imfinity/infinity which is undefined. using L'Hospital's Rule
d(e^4x)/dx=64e^4x
d(x^2)/dx=2x
4e^4x/2x=2e^4x/x
still infinity/infinity Use LoHospitals Rule again & get
8e^4x/1 lim x---->infinity becomes infinity.

Answer 4

you ought to stick to l'hopitals rule you had the ultimate suited physique of concepts: rewrite it as lim sin(a million/x) / (a million/x) now stick to lhopital via taking the derivative of top and backside (-cos(a million/x) / x²) / (-a million/x²) = cos(a million/x) now lim cos(a million/x) = cos (0) = a million

Answer 5

Lim[(e^(4x)-1-4x)/(x^2)]= (inf-inf)/inf
so use l'hopital's rule
Lim[(4e^(4x)-4)/(2x)]= (inf/inf)
so use l'hopital's rule
Lim[(16e^(4x))/(2)]= inf/2=inf

Answer 6

JUST DIVIDE BY ZERO!