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Babynini
 one year ago
Limits....again. Help!
Babynini
 one year ago
Limits....again. Help!

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Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Both numbers 6 and 7, but let's begin with 6!!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1L'hospital for first one

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Ya :) Imma wait and see what Ganeshie has to say though ^^

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5Notice that the denominator is 0 when you plugin x=0 Also, for the limit to exist (finite), you must get the indeterminate 0/0 form when you plugin x=0. otherwise the limit does not exist (infinite). so, the numerator part must evaluate to 0 at x=0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\(\sqrt{ax+b}2\) evaluating it at \(x=0\) and setting equal to \(0\) does give you \(b=4\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5plugin the \(b\) value, rationalize and you will see what \(a\) needs to be

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1For 7 note this dw:1444533414414:dw you can do the same with 2x1 you'll have an ugly numerator at the beginning, but it'll simplify to something nice :D

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Right, so I got b. But the a is being something terrible.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.5\[\frac{\sqrt{ax+4}2}{x}=\frac{ax+44}{x(\sqrt{ax+4}+2)} = \frac{a}{\sqrt{ax+4}+2}\] plugin x=0 and you will see immediately that \(a\) needs to be \(4\)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 thanks! I got it :)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 gwow you are doing some fancy stuff, I think I also follow wht you are doing :D

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0mm well if a = 4 and b = 4..then it works.

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0but it's always over 0 :/

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Ok, yeah that's what I got xD thanks yall :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.0you actully do get a limit!

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay i hope thats clear now

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i think this is how lhopital is justified too

dan815
 one year ago
Best ResponseYou've already chosen the best response.0we are looking at the behavior of your function very locally, so we'd only care about the slopes changes around that point

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Righto, 6 all makes sense :) I'm going to start a new thread for number 7

dan815
 one year ago
Best ResponseYou've already chosen the best response.0so for example the lhopital way d/x(abs(2x1) abs(2x+))  d/dx(x) d/dx ( abs(2x1)  d/dx(abs(2x+1))  1 now derivative of abs(2x1) near 0 = is 2 and the detiaive of abs(2x+1) near 0 is 2 so 2  2  1 =4 as x>0

Babynini
 one year ago
Best ResponseYou've already chosen the best response.0Em, haven't learnt derivatives yet :/

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay then do it the other way

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I think multiplying by conjugate might work
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