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hanifah
 one year 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
hanifah
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1Can we use L'hospital's rule?

hanifah
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1lhospital is working for me

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

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

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

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

myininaya
 one year 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.

hanifah
 one year 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
 one year 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.

Jonask
 one year ago
Best ResponseYou've already chosen the best response.2yes ...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

hanifah
 one year 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

Jonask
 one year ago
Best ResponseYou've already chosen the best response.2yes 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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