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anonymous
 one year ago
Limit as x approaches infinity of 8x[ln(x+9)  ln(x)]
anonymous
 one year ago
Limit as x approaches infinity of 8x[ln(x+9)  ln(x)]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How can I rewrite it as a fraction in order to apply L'Hopital's rule?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3you can bring x in the denominator and write it as 1/x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right but then plugging in infinity I would get infinity/0 right? And thats not defined

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3you wanted to apply LH right ? so first differentiate numerator and denominator separately later think about plugging it, after some simplification

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats 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
 one year ago
Best ResponseYou've already chosen the best response.3right, did you try the substitution x =1/y ? when x> infinity, y>0

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\left(\ln(x+9)\ln(x)\right)}\) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }\left(\ln\frac{x+9}{x}\right)}\) \(\large\color{slate}{\displaystyle\left(\ln\left(\lim_{x \rightarrow ~\infty }\frac{x+9}{x}\right)\right)}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0so you are taking the ln of that limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes thats right! Oh but where's the 8x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And so is that also one of the limit properties? Exchanging the places of ln and limit?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0yes limit properties allow thatr

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright so using the same idea where would the 8x go?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3its not exchanging... Limit can be dragged inside of only those functions which are continuous

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3can you give the substitution one shot? the substitution x =1/y ? when x> infinity, y>0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh now that makes sense! I have to review those properties

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would I apply that method @hartnn

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3plug 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
 one year ago
Best ResponseYou've already chosen the best response.0Okay I understand the steps.. just not clear on the concept, why are we saying that x=1/y?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0limit change is fine

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0if not the only...

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3its a substitution, it may/may not make the solution easier/faster to find... just a try

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3can you try to combine 2 logs? \(\ln (1/y+9)  \ln (1/y)\)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3the limit really becomes simple after that...you'll easily be able to apply LH rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That would be ln[ (1/y +9)/(1/y)]

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3yep, simplify that further

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it simplifies to ... yah exactly haha! I was just about to type that in

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3good! now is your new limit in the form of 0/0? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey umm sorry to interrupt but could u help me pleeaassee!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But we no longer have a fraction, do we?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3remember the 8x became 8/y?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the we have (8ln(1+9y))/y which is equal to 0/0 ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0if it's 0/0 that means you can use LH rule again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alrighty! Yeah exactly!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0When you don't have the 0/0 form anymore, you can't use the rule.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So now to differentiate, do I use implicit diff ?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3Implicit? you mean chain rule for numerator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Or do i just change the y's to 1/x again?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{(8\ln(1+9y))}{y}\] this?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3thats the beauty of substitution, you don't need to change back :)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3whatever answer you get for that limit, will be the answer for your original limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im just thinking of the y's? do i diff with respect to y then?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3yes, diff. w.r.t y only

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh ok that makes sense!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0product rule on the top?

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3there are no productions on the top? chain rule... its like a function inside a function linear function inside log function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0theres also an 8y tho

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh no the y's on the bottom haha sorry!

hartnn
 one year ago
Best ResponseYou've already chosen the best response.38y ?? there is only 8 in the numerator, right ? 8 is a constant and can be brought out.. lol ok

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so I'm hoping i got it right in the end... is it 8/1+9y

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0@hartnn constants don't have derivatives... it goes to 0 Ex. the derivative of 6 is 0

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3not exactly \([\ln (1+9y)]' = (1/(1+9y))(1+9y)'\) chain rule you missed differentiating 1+9y

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3thats right Usuki, here the constant 8 is multiplied with log function its not alone thats why we can bring it out of limit

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3yep now you can just plug in y= 0 :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh so that leaves us with 8*9 = 72? :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hooray!!! Thats correct! :D

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3congrats! and welcome ^_^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much!! I was getting totally hopeless on this one xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks to everyone who helped out! :)
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