Question

Just a couple seconds tutoring on limits, something I got confused about.

Answer 1

\[\huge\color{blue}{ \lim_{x \rightarrow ∞} ~~\frac{2x^2-5}{x(x+4)} } \]

I saw the answer was 2, but didn't quite get why.
I foiled the bottom and got
\[\huge\color{blue}{ \lim_{x \rightarrow ∞} ~~\frac{2x^2-5}{x^2+4x} } \]

then using L'H'S I get
\[\huge\color{blue}{ \lim_{x \rightarrow ∞} ~~\frac{(2x^2-5)'}{(x^2+4x)'} } \]
\[\huge\color{blue}{ \lim_{x \rightarrow ∞} ~~\frac{4x-0}{2x+4} } \]
\[\huge\color{blue}{ \lim_{x \rightarrow ∞} ~~\frac{2x}{x+2} } \]
so the final answer would then be 0/2 or zero, not 2, right?

Answer 2

no it would be 2 ,
\[\frac{ (2x)' }{ (x+2)' }=\frac{ 2 }{ 1 }\]

Answer 3

deriving it a second time, after having done it once already?

Answer 4

yeah , as long as you have limit of form inf/inf you can do it

Answer 5

So doing it the standard way, deriving once and plugging in 0, is not correct?
I thought L'H'S is derive top and bottom and then plug in whatever x is approaching to.

Answer 6

I never thought then in inf limit you can derive it more than once.

Answer 7

why you plug in 0 if limit goes to infinity

Answer 8

oh, yeah, didn't think abt that, my bad; ty

Answer 9

I see, so using L'H'S in inf limits I'll just keep deriving until I get a constant (or a number without a variable)

Answer 10

yeah u can derive as many time as you want until the condition of L'H'S is satisfied

Answer 11

Ok, ty!

Answer 12

ur welcome

Answer 13

Review L'Hospital's rule once more.

Answer 14

I am pretty sure I did that just now, ty for your suggestion, Isaiah.