Question

L'hopitals rule again :(

Answer 1

\[\lim_{x \rightarrow oo}\sqrt{x^2+7x+4}-x\]

Answer 2

@jim_thompson5910

Answer 3

intuition tells me infinity

Answer 4

wrong..

Answer 5

no the -x is on the outside

Answer 6

one sec uploading solution

Answer 7

rationalize first then you'll get your equation in the form that it would be easy to apply the rule

Answer 8

wow

Answer 9

?

Answer 10

ya i did start by rationalizing but it wasnt that obvious

Answer 11

yeah the hardest l'hopital's problems are usually the ones that involve indeterminate forms inf-inf and 0*inf

Answer 12

the trick is to use exponents, logarithms, or fractions to otherwise get you where you need to be

Answer 13

the wolfram solution didnt even use l'hopitals but this seems to be the only way id know to do it.

Answer 14

thanks..

Answer 15

@TylerD starting from step two after rationalization you can use l'hopital's

Answer 16

because you have form infinity/infinity

Answer 17

ud have \[\frac{ 7 }{ \frac{ 2x+7 }{ 2\sqrt{x^2+7x+4} }+1 }\]

Answer 18

or something ridic like that but

Answer 19

do u wanna know where lhopital rule comes from

Answer 20

sure why not :)

Answer 21

u think the world just gives u everything on a silver platter

Answer 22

?

Answer 23

https://www.youtube.com/watch?v=SIB4WDYF5DQ

Answer 24

it comes from taylor's magic hat

Answer 25

does the world give me silver platters on silver platters?

Answer 26

yes :)

Answer 27

look at that proof for special case its nice, it comes comes first principles

Answer 28

f(0)=4 and g(0)=2?

Answer 29

I think rationalizing works pretty neatly as suggested by @canni_fanni earlier... I don't see an easy way to force it to apply Lhopital..

Answer 30

so you cant actually use l'hopitals rule on this problem?

Answer 31

ofcourse you can use
all i am saying is that I don't see how it simplifies matters compared to rationalizing here

Answer 32

are u saying the taylor serise of
f/g = taylor series of f'/g' if f(a) and g(a) =0

Answer 33

damn typoes :/

short proof of Lhopital rule :

By taylor series,
\[\lim\limits_{x\to a}\dfrac{f(x)}{g(x)} = \lim\limits_{x\to a}\dfrac{f(a) + (x-a)f'(a) + \dfrac{1}{2!}(x-a)^2f''(a)+ \cdots }{g(a) + (x-a)g'(a) + \dfrac{1}{2!}(x-a)^2g''(a)+ \cdots}\]

which equals \(\dfrac{f'(a)}{g'(a)}\) if \(f(a) = g(a) = 0\)

Answer 34

Looks numerator and denominator are getting separate treatment.
I have just expanded numerator and denominator of fraction

Answer 35

more steps :
\[\begin{align}\lim\limits_{x\to a}\dfrac{f(x)}{g(x)} &= \lim\limits_{x\to a}\dfrac{f(a) + (x-a)f'(a) + \dfrac{1}{2!}(x-a)^2f''(a)+ \cdots }{g(a) + (x-a)g'(a) + \dfrac{1}{2!}(x-a)^2g''(a)+ \cdots}\\~\\~\\
&= \lim\limits_{x\to a}\dfrac{0 + (x-a)f'(a) + \dfrac{1}{2!}(x-a)^2f''(a)+ \cdots }{0 + (x-a)g'(a) + \dfrac{1}{2!}(x-a)^2g''(a)+ \cdots}\\~\\~\\
&= \lim\limits_{x\to a}\dfrac{f'(a) + \dfrac{1}{2!}(x-a)f''(a)+ \cdots }{g'(a) + \dfrac{1}{2!}(x-a)g''(a)+ \cdots}\\~\\~\\
&=\dfrac{f'(a)}{g'(a)}\\~\\
&= \lim\limits_{x\to a}\dfrac{f'(x)}{g'(x)}\\~\\
\end{align}\]

Answer 36

we can apply lhopital again if we again have \(f'(a) = g'(a) = 0\)