How do I do this limit?( without hopital)

Question

limx-->positive infty sqrt(9x^3+x) -x^3/2) 

 

The only one thats under the square root is 9x^3+x

 

I dont know how to solve is

Answer 1

Note that sqrt(9x^3+x) - x^(3/2) > sqrt(9x^3) - x^(3/2) for all x > 0

 

Now note that sqrt(9x^3) - x^(3/2) = 3x^(3/2) - x^(3/2) = 2x^(3/2)

 

Since 2x^(3/2) clearly goes to infinity as x does, then so does sqrt(9x^3+x) - x^(3/2) (since it is greater than 2x^(3/2)).

Answer 2

A simpler problem like this one would be

 

lim x-->+infinity sqrt{9x²+x} - x

 

This could be solved by conjugating:

 

(9x²+x-x²) / (sqrt{9x²+x}+x)

 

Divide through numerator and denominator by the highest power of x,  x² , to get

 

(8 + 1/x) / (sqrt{9/x²+1/x³} + 1/x)

 

This approaches 8 / +0, hence lim = +infinity

 

In your problem, do the same thing