# Limit Help with absolute value in the denominator

• Aug 25th 2012, 10:49 AM
skinsdomination09
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)
• Aug 25th 2012, 11:04 AM
Plato
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 $x\to\infty$ that mean $x>0$. THUS $7-|x|=7-x$.
• Aug 25th 2012, 07:04 PM
Prove It
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 \begin{align*} \frac{\sqrt{4x^2 - 1}}{7} - |x| \end{align*}

or

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

or

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

or

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

or something else entirely...