# Thread: limit as x goes to infinity

1. ## limit as x goes to infinity

Consider f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8)
Take steps to find the limit as x -> infinity and –infinity
a) Find the limit of f(x) as x goes to infinity
rewrite the expression sqrt(9x**2 + 6) by extracting x**2 as follows:
sqrt(9x**2 + 6) = sqrt(x**2 ) sqrt(9x**2 + 6) = Ax sqrt(p(x))
p(x) = ________________ and A = ____ (constant)
b) It follows that f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8) = (A’ sqrt(p(x)) / (-6-8/x)
find p(x) and A’
c) what is the limit?
d) Now find as x goes to – infinity (exact same information, but with – infinity.)

my work has gotten some of these (in class we didn't use the same format webwork expects..)

f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8)
i took the sqrt(x**2 ) out and got (-4 sqrt(9+ 6/x**2) / (-6-8/x)
For the constant part of filling it in, -4 doesn't work for "A" and neither does - 12 (thought mauybe -4 * sqrt9)
anyway i found the limit as x goes to infinity which is 2.

considering i don't know what they want for the value of A, which is a constant, i don't know what they want for A'
I tired using teh value 0, (derivative of a constant) but that is wrong!

hope somoene can help me out on this A thing they're wanting of me!!!
thankyou
Brittany

2. Originally Posted by williamb
Consider f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8)
Take steps to find the limit as x -> infinity and –infinity
a) Find the limit of f(x) as x goes to infinity
rewrite the expression sqrt(9x**2 + 6) by extracting x**2 as follows:
sqrt(9x**2 + 6) = sqrt(x**2 ) sqrt(9x**2 + 6) = Ax sqrt(p(x))
p(x) = ________________ and A = ____ (constant)
b) It follows that f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8) = (A’ sqrt(p(x)) / (-6-8/x)
find p(x) and A’
c) what is the limit?
d) Now find as x goes to – infinity (exact same information, but with – infinity.)

my work has gotten some of these (in class we didn't use the same format webwork expects..)

f(x) = (-4 sqrt(9x**2 + 6) / (-6x-8)
i took the sqrt(x**2 ) out and got (-4 sqrt(9+ 6/x**2) / (-6-8/x)
For the constant part of filling it in, -4 doesn't work for "A" and neither does - 12 (thought mauybe -4 * sqrt9)
anyway i found the limit as x goes to infinity which is 2.

considering i don't know what they want for the value of A, which is a constant, i don't know what they want for A'
I tired using teh value 0, (derivative of a constant) but that is wrong!

hope somoene can help me out on this A thing they're wanting of me!!!
thankyou
Brittany

They ask you to write $\sqrt{9x^2 + 6}$ in the form $Ax \sqrt{p(x)}$. So $A = 1$ and $p(x) = 9 + \frac{6}{x^2}$.

So $f(x) = \frac{-4 \sqrt{9x^2 + 6}}{-6x - 8} = \frac{-4 \sqrt{9 + \frac{6}{x^2}}}{-6 - \frac{8}{x}}$. Compare this with $\frac{A' \sqrt{p(x)}}{-6 - \frac{8}{x}}$ and it's clear that $A' = -4$.

### proof of limit f(x) goes to - infinity as x goes to - infinity

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