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anonymous
 3 years ago
if a>0 what is the
lim(x>a) ((sqrt(2xa^3  x^4))  a(cuberoot((a^2)(x))))/(a(ax^3)^(1/4))
What method can help simplify this better? I started by multiplying top and bottom by the conjugate and it was still pretty messy. equation attached in comments for clarity. Any help would be appreciated! thanks
anonymous
 3 years ago
if a>0 what is the lim(x>a) ((sqrt(2xa^3  x^4))  a(cuberoot((a^2)(x))))/(a(ax^3)^(1/4)) What method can help simplify this better? I started by multiplying top and bottom by the conjugate and it was still pretty messy. equation attached in comments for clarity. Any help would be appreciated! thanks

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myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Can we use L'hospital's rule?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, I tried that first and got a very long/algebraic nightmare.. is it possible to multiply by the conjugate and then us lhr?

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1lhospital is working for me

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1all you need is one round and then plug in

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1remember to treat a as constant

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0using L'hospital's rule \[\Large \lim _{x\to a}\frac{\frac{2a^34x^3}{2(2ax^3x^4)^\frac{1}{2}}a\frac{a^2}{3(a^2x)^\frac {2}{3}}}{\frac{3ax^2}{4(ax^3)^\frac{3}{4}}}=\frac{a\frac{a}{3}}{\frac{3a^3}{4a^3}}=a\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahhh, so I got the derivatives correct, but I simplified to something that is i don't even know.. haha. honest mistake that will probably kill me on the final=/.. giving the medal to @Jonask this round sorry @myininaya i appreciate the help though!!

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.14/3 divided by 3/4 is 16/9

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1Also I didn't give you the answer because I wanted you to do it yourself. lol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know.. @myininaya i'm guessing you're a teacher? haha joking.. jonask doesn't have a medal also I'm not entirely sure what the rules of openstudy are.. i just ask questions, but i feel like some people take the medals pretty seriously. I will sleep on it and get back to you :P

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1No I don't care about the medals. I just want the students to be able to get the answers themselves. This isn't suppose to be a site when you can come to cheat.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes ...the answer shud be 16a/9...since this was a messy job i decided to write the solution wich u obviously attemted...hence u cud see your mistake

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0agree, i couldn't figure out where i messed up via wolfram so open study is the next best thing. i need to improve my algebra

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes you are write,thats a good idea ,wolfram can help you only if you know ur algebra already,u jus need it for verification and visualisation...we should sharpen ourselves first before we use machines
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