Limit Help with absolute value in the denominator

**what is **

the limit as x--> - infinity of √(4x²-1)/ 7 - lxl

I don't understand how to treat absolute value of x in the denominator.

I assume this would be indetirminate as you end up with neg inf./inf

so I would have to take the derivitive (l'hospital's rule) of the top and bottom

the top would become 4x ( 4x²-1) to the -1/2 power

the bottom would become X+1/ lxl??

I'm really unsure and I feel like this may not exist.

Thanks for all your help, you guys are great!!!!!!!!!!!! (Wink)

(Wink)(Wink)(Wink)(ps. **no calc. question)**

Re: Limit Help with absolute value in the denominator

Quote:

Originally Posted by

**skinsdomination09** what is

the limit as x--> - infinity of √(4x²-1)/ 7 - lxl[/B]

As $\displaystyle x\to\infty$ that mean $\displaystyle x>0$. THUS $\displaystyle 7-|x|=7-x$.

Re: Limit Help with absolute value in the denominator

Quote:

Originally Posted by

**skinsdomination09** **what is **

the limit as x--> - infinity of √(4x²-1)/ 7 - lxl

I don't understand how to treat absolute value of x in the denominator.

I assume this would be indetirminate as you end up with neg inf./inf

so I would have to take the derivitive (l'hospital's rule) of the top and bottom

the top would become 4x ( 4x²-1) to the -1/2 power

the bottom would become X+1/ lxl??

I'm really unsure and I feel like this may not exist.

Thanks for all your help, you guys are great!!!!!!!!!!!! (Wink)

(Wink)(Wink)(Wink)(ps. **no calc. question)**

You need to learn some LaTeX, or else use brackets where they're needed. I can't tell if this is

$\displaystyle \displaystyle \begin{align*} \frac{\sqrt{4x^2 - 1}}{7} - |x| \end{align*}$

or

$\displaystyle \displaystyle \begin{align*} \frac{\sqrt{4x^2 - 1}}{7 - |x|} \end{align*}$

or

$\displaystyle \displaystyle \begin{align*} \sqrt{\frac{4x^2 - 1}{7} - |x|} \end{align*}$

or

$\displaystyle \displaystyle \begin{align*} \sqrt{\frac{4x^2 - 1}{7 - |x|}} \end{align*}$

or something else entirely...