if a0 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

Question

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 · Answer

attachment

myininaya · Answer

Can we use L'hospital's rule?

anonymous · Answer

Yeah, I tried that first and got a very long/algebraic nightmare.. is it possible to multiply by the conjugate and then us lhr?

myininaya · Answer

lhospital is working for me

myininaya · Answer

all you need is one round and then plug in

myininaya · Answer

remember to treat a as constant

anonymous · Answer

using L'hospital's rule
$$\Large \lim _{x\to a}\frac{\frac{2a^3-4x^3}{2(2ax^3-x^4)^\frac{1}{2}}-a\frac{a^2}{3(a^2x)^\frac {2}{3}}}{-\frac{3ax^2}{4(ax^3)^\frac{3}{4}}}=\frac{-a-\frac{a}{3}}{\frac{-3a^3}{4a^3}}=a$$

anonymous · Answer

ahhh, 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 · Answer

-4/3 divided by -3/4 is 16/9

myininaya · Answer

Also I didn't give you the answer because I wanted you to do it yourself. lol.

anonymous · Answer

i 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 · Answer

No 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 · Answer

yes ...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 · Answer

agree, 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 · Answer

yes 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