Question

Find the limit as x approaches infinity
of sqrt(x^4+4x^3+5) / 10x^2-8

Answer 1

hint : the numerator and the denominator divided by x^2

Answer 2

at @RadEn so I did that and I can see that if I look at the exponents and since its x^6/x^4 I would get my limit to be 1/10 ?

Answer 3

the answer is right, it is 1/10. but i dont knw where did you get x^6/x^4 ?

Answer 4

Well the x^6/x^4 was just part of my answer, sorry I did not type my work. I hadn't thought of dividing by x^2, so thank you so much for your hint :)

Answer 5

actually, if the numerator is divided by x^2, get
sqrt(x^4+4x^3+5)/x^2 = sqrt(x^4+4x^3+5)/sqrt(x^4) = sqrt((x^4+4x^3+5)/x^4) = sqrt(1+4/x+5/x^4)

and if the denominator divided by x^2, get
(10x^2-8 )/x^2 = 10 - 8/x^2

so, the limit becomes
limit x->inf sqrt(1+4/x+5/x^4)/(10 - 8/x^2) = 1/10

Answer 6

Oh I just looked at my work and I did some mistakes but now I understand it better, thank you.

Answer 7

you're welcome