anonymous
  • anonymous
Limit as x approaches infinity of 8x[ln(x+9) - ln(x)]
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
How can I rewrite it as a fraction in order to apply L'Hopital's rule?
hartnn
  • hartnn
you can bring x in the denominator and write it as 1/x
hartnn
  • hartnn
|dw:1435026931030:dw|

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anonymous
  • anonymous
Right but then plugging in infinity I would get infinity/0 right? And thats not defined
hartnn
  • hartnn
you wanted to apply LH right ? so first differentiate numerator and denominator separately later think about plugging it, after some simplification
anonymous
  • 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
  • hartnn
right, did you try the substitution x =1/y ? when x-> infinity, y->0
SolomonZelman
  • 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
  • SolomonZelman
so you are taking the ln of that limit
SolomonZelman
  • SolomonZelman
0
SolomonZelman
  • SolomonZelman
oh 8x !!1soprry
anonymous
  • anonymous
Yes thats right! Oh but where's the 8x?
SolomonZelman
  • SolomonZelman
yes yes, my bad
anonymous
  • anonymous
And so is that also one of the limit properties? Exchanging the places of ln and limit?
SolomonZelman
  • SolomonZelman
yes limit properties allow thatr
anonymous
  • anonymous
Alright so using the same idea where would the 8x go?
hartnn
  • hartnn
its not exchanging... Limit can be dragged inside of only those functions which are continuous
hartnn
  • hartnn
can you give the substitution one shot? the substitution x =1/y ? when x-> infinity, y->0
anonymous
  • anonymous
Ohh now that makes sense! I have to review those properties
anonymous
  • anonymous
How would I apply that method @hartnn
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
  • anonymous
Okay I understand the steps.. just not clear on the concept, why are we saying that x=1/y?
SolomonZelman
  • SolomonZelman
limit change is fine
SolomonZelman
  • SolomonZelman
if not the only...
hartnn
  • hartnn
its a substitution, it may/may not make the solution easier/faster to find... just a try
anonymous
  • anonymous
Ohh ok
hartnn
  • hartnn
can you try to combine 2 logs? \(\ln (1/y+9) - \ln (1/y)\)
hartnn
  • hartnn
ln A - ln B = ln A/B
hartnn
  • hartnn
the limit really becomes simple after that...you'll easily be able to apply LH rule
anonymous
  • anonymous
That would be ln[ (1/y +9)/(1/y)]
hartnn
  • hartnn
yep, simplify that further
hartnn
  • hartnn
|dw:1435028541306:dw|
anonymous
  • anonymous
So it simplifies to ... yah exactly haha! I was just about to type that in
hartnn
  • hartnn
good! now is your new limit in the form of 0/0? :)
anonymous
  • anonymous
hey umm sorry to interrupt but could u help me pleeaassee!
anonymous
  • anonymous
its urgent
anonymous
  • anonymous
But we no longer have a fraction, do we?
hartnn
  • hartnn
remember the 8x became 8/y?
anonymous
  • anonymous
Oh yeah!!
anonymous
  • anonymous
so the we have (8ln(1+9y))/y which is equal to 0/0 ?
anonymous
  • anonymous
then*
hartnn
  • hartnn
yes :3
UsukiDoll
  • UsukiDoll
if it's 0/0 that means you can use LH rule again
anonymous
  • anonymous
Alrighty! Yeah exactly!
UsukiDoll
  • UsukiDoll
When you don't have the 0/0 form anymore, you can't use the rule.
hartnn
  • hartnn
yaay :D
anonymous
  • anonymous
So now to differentiate, do I use implicit diff ?
hartnn
  • hartnn
Implicit? you mean chain rule for numerator?
anonymous
  • anonymous
Or do i just change the y's to 1/x again?
UsukiDoll
  • UsukiDoll
\[\frac{(8\ln(1+9y))}{y}\] this?
hartnn
  • hartnn
thats the beauty of substitution, you don't need to change back :)
hartnn
  • hartnn
whatever answer you get for that limit, will be the answer for your original limit
anonymous
  • anonymous
Im just thinking of the y's? do i diff with respect to y then?
hartnn
  • hartnn
yes, diff. w.r.t y only
anonymous
  • anonymous
Ohh ok that makes sense!
anonymous
  • anonymous
product rule on the top?
hartnn
  • hartnn
there are no productions on the top? chain rule... its like a function inside a function linear function inside log function
anonymous
  • anonymous
theres also an 8y tho
anonymous
  • anonymous
oh no the y's on the bottom haha sorry!
hartnn
  • hartnn
8y ?? there is only 8 in the numerator, right ? 8 is a constant and can be brought out.. lol ok
anonymous
  • anonymous
okay so I'm hoping i got it right in the end... is it 8/1+9y
UsukiDoll
  • UsukiDoll
@hartnn constants don't have derivatives... it goes to 0 Ex. the derivative of 6 is 0
hartnn
  • hartnn
not exactly \([\ln (1+9y)]' = (1/(1+9y))(1+9y)'\) chain rule you missed differentiating 1+9y
hartnn
  • 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
  • anonymous
8*9 / 1+9y ?
hartnn
  • hartnn
yep now you can just plug in y= 0 :)
anonymous
  • anonymous
Ahhh so that leaves us with 8*9 = 72? :O
hartnn
  • hartnn
\(\huge \checkmark \)
anonymous
  • anonymous
Hooray!!! Thats correct! :D
hartnn
  • hartnn
congrats! and welcome ^_^
anonymous
  • anonymous
Thanks so much!! I was getting totally hopeless on this one xD
anonymous
  • anonymous
Thanks to everyone who helped out! :)

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