Limit as x approaches infinity of 8x[ln(x+9) - ln(x)]

- anonymous

Limit as x approaches infinity of 8x[ln(x+9) - ln(x)]

- jamiebookeater

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- anonymous

How can I rewrite it as a fraction in order to apply L'Hopital's rule?

- hartnn

you can bring x in the denominator and write it as 1/x

- hartnn

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## More answers

- anonymous

Right but then plugging in infinity I would get infinity/0 right? And thats not defined

- hartnn

you wanted to apply LH right ?
so first differentiate numerator and denominator separately
later think about plugging it, after some simplification

- anonymous

Thats true but in order to use LH we have to first make sure that we have an indeterminate form and oo/0 isn't it

- hartnn

right,
did you try the substitution
x =1/y ?
when x-> infinity,
y->0

- SolomonZelman

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\left(\ln(x+9)-\ln(x)\right)}\)
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\left(\ln\frac{x+9}{x}\right)}\)
\(\large\color{slate}{\displaystyle\left(\ln\left(\lim_{x \rightarrow ~\infty }\frac{x+9}{x}\right)\right)}\)

- SolomonZelman

so you are taking the ln of that limit

- SolomonZelman

0

- SolomonZelman

oh 8x !!1soprry

- anonymous

Yes thats right! Oh but where's the 8x?

- SolomonZelman

yes yes, my bad

- anonymous

And so is that also one of the limit properties? Exchanging the places of ln and limit?

- SolomonZelman

yes limit properties allow thatr

- anonymous

Alright so using the same idea where would the 8x go?

- hartnn

its not exchanging...
Limit can be dragged inside of only those functions which are continuous

- hartnn

can you give the substitution one shot?
the substitution
x =1/y ?
when x-> infinity,
y->0

- anonymous

Ohh now that makes sense! I have to review those properties

- anonymous

How would I apply that method @hartnn

- hartnn

plug in x= 1/y
so y = 1/x -> 1/infinity->0
8x becomes 8/y
ln (x+9) becomes ln (1/y + 9)
ln x becomes ln (1/y) or

- anonymous

Okay I understand the steps.. just not clear on the concept, why are we saying that x=1/y?

- SolomonZelman

limit change is fine

- SolomonZelman

if not the only...

- hartnn

its a substitution,
it may/may not make the solution easier/faster to find...
just a try

- anonymous

Ohh ok

- hartnn

can you try to combine 2 logs?
\(\ln (1/y+9) - \ln (1/y)\)

- hartnn

ln A - ln B = ln A/B

- hartnn

the limit really becomes simple after that...you'll easily be able to apply LH rule

- anonymous

That would be ln[ (1/y +9)/(1/y)]

- hartnn

yep,
simplify that further

- hartnn

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- anonymous

So it simplifies to ... yah exactly haha! I was just about to type that in

- hartnn

good!
now is your new limit in the form of 0/0? :)

- anonymous

hey umm sorry to interrupt but could u help me pleeaassee!

- anonymous

its urgent

- anonymous

But we no longer have a fraction, do we?

- hartnn

remember the 8x became 8/y?

- anonymous

Oh yeah!!

- anonymous

so the we have (8ln(1+9y))/y which is equal to 0/0 ?

- anonymous

then*

- hartnn

yes :3

- UsukiDoll

if it's 0/0 that means you can use LH rule again

- anonymous

Alrighty! Yeah exactly!

- UsukiDoll

When you don't have the 0/0 form anymore, you can't use the rule.

- hartnn

yaay :D

- anonymous

So now to differentiate, do I use implicit diff ?

- hartnn

Implicit?
you mean chain rule for numerator?

- anonymous

Or do i just change the y's to 1/x again?

- UsukiDoll

\[\frac{(8\ln(1+9y))}{y}\] this?

- hartnn

thats the beauty of substitution,
you don't need to change back :)

- hartnn

whatever answer you get for that limit, will be the answer for your original limit

- anonymous

Im just thinking of the y's? do i diff with respect to y then?

- hartnn

yes, diff. w.r.t y only

- anonymous

Ohh ok that makes sense!

- anonymous

product rule on the top?

- hartnn

there are no productions on the top?
chain rule...
its like a function inside a function
linear function inside log function

- anonymous

theres also an 8y tho

- anonymous

oh no the y's on the bottom haha sorry!

- hartnn

8y ??
there is only 8 in the numerator, right ?
8 is a constant and can be brought out..
lol ok

- anonymous

okay so I'm hoping i got it right in the end... is it 8/1+9y

- UsukiDoll

@hartnn constants don't have derivatives... it goes to 0
Ex. the derivative of 6 is 0

- hartnn

not exactly
\([\ln (1+9y)]' = (1/(1+9y))(1+9y)'\)
chain rule
you missed differentiating 1+9y

- hartnn

thats right Usuki,
here the constant 8 is multiplied with log function
its not alone
thats why we can bring it out of limit

- anonymous

8*9 / 1+9y ?

- hartnn

yep
now you can just plug in y= 0 :)

- anonymous

Ahhh so that leaves us with 8*9 = 72? :O

- hartnn

\(\huge \checkmark \)

- anonymous

Hooray!!! Thats correct! :D

- hartnn

congrats!
and welcome ^_^

- anonymous

Thanks so much!! I was getting totally hopeless on this one xD

- anonymous

Thanks to everyone who helped out! :)

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