Question

Limit question

Answer 1

Lim x approaches negative infinity \[\frac{ 16x }{ \sqrt{16x^2 -9} }\]

Answer 2

Would the answer be 4? Or negative 4?

Answer 3

We're approaching negative infinity,\[\large\rm \frac{-16x}{+\sqrt{16x^2-9}}\]Think of it like this maybe? :o
The numerator is going to spit out a big negative number,
the denominator is going to give us positive always.

Answer 4

That makes sense! So it will be -4 Zep?

Answer 5

@zepdrix quick question, would lhopital's rule work for this?

Answer 6

lhopital's rule is actually what we're learning this unit! I tried using it but couldn't figure it out..

Answer 7

-4? Yes.

L'Hopital's Rule? Hmm interesting :D
Ya maybe... it's giving us the indeterminate form \(\large\rm \frac{-\infty}{\infty}\).
I don't think the signs have to match for L'Hop...

Answer 8

Oh wow, L'Hopital actually doesn't work out because of how the square root derivative turns out.\[\rm \lim_{x\to-\infty}\frac{16x}{\sqrt{16x^2-9}}\quad L'H=\quad \lim_{x\to-\infty}\frac{16}{\frac{16x}{\sqrt{16x^2-9}}}\quad=\quad \lim_{x\to-\infty}\frac{\sqrt{16x^2-9}}{x}\]Which is now giving us the indeterminate form \(\large\rm \frac{\infty}{-\infty}\).
And if we apply L'Hop again we end up with the same function we started with! Hah!

Answer 9

Interesting. That's what threw me off initially, I did this problem on a quiz and tried using L'Hops. Thanks for making this clear zebbles!

Answer 10

yay team