Find the limit or show that it does exist, lim goes to infinity of (sqrt 9x^2+x-3x)

Question

Help me please

Answer 1

I think you meant   √(9x² + x - 3).

 

 

When x approaches infinity, we expect f(x) to get closer to a specific value, or f(x) increases infinitely or decreases infinitely.  Horizontal asymptotes are usually found using this concept.  So lets pick values of x to that we can evaluate the function.  Since the number under the square-root cannot be negative, we need to find values in which the function is defined.

 

9x² + x - 3 ≥ 0

 

If our zeros are

 

x = (-1 ± √(1 - 4(-27))) / 18

x = (-1 ± 10.44) / 18

x =  (-1 ± 10.44) / 18           and             x =  (-1 ± 10.44) / 18

 

x = -0.636                            and             x = 0.524

 

 

f(-1) = √(9 - 1 + 3)

        = √(11)

 

f(0) = √(-3)

 

f(1) = √(9 + 1 - 3)

      = √(7)

 

Then the domain is  (-∞, 0)∪(0, ∞).

 

Evaluate f(x) when

 

x = 1

x = 10

x = 100

x = 10000

x = 1000000

x = 1000000000

 

If f(x) does not get closer to a constant value as you evaluate it, then the limit does not exist.