what is the limit of f(x)...?

Question

if x -> infinity and f(x)=(square root of (9x^6 - x) ) / (x^3 +1)

Explain please cuz i know what the answer is, i wanna know y thats the answer....:)thx

ummm...the answer's 3, but y?

Answer 1

you don't need to use L'Hospital rule. Just use some algebra

f(x) = √(9x^6 - x) / (x^3 + 1)

rewrite the expression
          √(9x^6 - x)
f(x) = ---------------
          √(x^3 + 1)²

expand

          √(9x^6 - x)
Lim = -----------------------
          √ (x^6 + 2x^3 + 1)
x -> infinity

now divide all the terms by x^6 for both top and bottom, and you'll get

          √(9 + 1/x^5)
Lim = --------------------
          √(1 + 2/x^3 + 1/x^6)
x -> infinity

as x approaches infinity:
1/x^5 approaches 0
2/x^3 approeaches 0
1/x^6 approaches 0

          √(9 + 0)
Lim = -----------------
          √(1 + 0 + 0)
x -> infinity

so all you have left is:
Lim = √(9) / √(1)
lim x -> infinity

which is 3

hope it helps.

Answer 2

Divide both the numerator and denominator of the fraction by x^3. Then use the property that c/x^n = 0 as x approaches ∞, where c is any constant and n>0

Answer 3

sqrt(9x^6 - x) / (x^3 + 1) use Ho'pital's rule (derive top and bottom) for instances such as 0/0 or inf / inf