L'Hôpital's question with limit

the question with the working

Do it again. Youd need to differeniate the denominator twice to get a non-zero value when x=0, so you must do the same to the numerator. If youve done Taylor series, it must mean the both the numerator and denominator have a Taylor/maclaurin series of the form
0 + 0x + cx^2/2 + ...
and c=1 for the numertor and c=2 for the denominator. You need to differenitate twice to get to the first nonzero value when x=0.

i wil differentiate again and send you what happens, no no technically we dont know taylor series so i. cant use them

You dont have to use the series, its just another way of thinking about what happens when you use l'hopital.

ive done it and found the b=1
how will i find a? since when i applies the law twice the a does not exist anymore so how will i find it?

ive done it and found the b=1
how will i find a? since when i applies the law twice the a does not exist anymore so how will i find it?

what must the value be for the first derivative (first application of l'hopital) to be zero?

why zero? isnt it 1/2
and how do i get rid of the limit part

This is one of the things where thinking about it as a series helps, but using l'hopital throughout, you know from the denominator that you'll have to do l'hopital (differentiate and take the limit) twice to end up with the first non-zero value. That means you must do the same to the numerator as the limit for the function is finite. So the first application l'hopital must give 0/0 which is indeterminate so you do it again. So the numerator for the first applicaiton of l'hopital is
0 = 1 + A
so ...

why 0 = 1 + A i dont understand why you did that and where it came from
i also underlined some things you said and that i did not understand

When you do the first applicaiton of l'hopital on the numerator, what do you get?

>> That means you must do the same to the numerator as the limit for the function is finite.

A fuller description of l'hopital is
https://www.mathsisfun.com/calculus/l-hopitals-rule.html
or your textbook or ... But for this question youre told the limit of
n(x)/d(x)
is finite and equal to 1/2. The first application of l'hopital will produce
n'(0) / 0
If n'(0) is not zero, then the limit of the original function is not finite. The only way the limit of the original function is finite is if n'(0)=0 and therefore the first applicaiton of l'hopital is indeterminant and therefore you have to do l'hopital again to get
n''(0) / d''(0)
We know d''(0) = 2, so n''(0) must equal 1.

>>So the numerator for the first applicaiton of l'hopital is 0 = 1 + A

Hopefully the above covers this now?

a little bit better i will look over these again llater..
also what does finite mean again?

1/0 is not finite. You could imagine it as +/-infinity or more properly say its indeterminant. 1/2 is finite.

When you do the first applicaiton of l'hopital on the numerator, what do you get?

>> That means you must do the same to the numerator as the limit for the function is finite.

A fuller description of l'hopital is
https://www.mathsisfun.com/calculus/l-hopitals-rule.html
or your textbook or ... But for this question youre told the limit of
n(x)/d(x)
is finite and equal to 1/2. The first application of l'hopital will produce
n'(0) / 0
If n'(0) is not zero, then the limit of the original function is not finite. The only way the limit of the original function is finite is if n'(0)=0 and therefore the first applicaiton of l'hopital is indeterminant and therefore you have to do l'hopital again to get
n''(0) / d''(0)
We know d''(0) = 2, so n''(0) must equal 1.

>>So the numerator for the first applicaiton of l'hopital is 0 = 1 + A

Hopefully the above covers this now?

i understand how the rule went and why we had to do it twice but i did not know you could seperately set equal 2x=2 the denominator since there is still the limit can i do that?? and also if i do that as you said ( but please verify if i can do this ie put the denominators equal 2x=2 even though there is a limit still)
i get x=1 but why do i put the numerator equal to 0?? the numerator is equal to 1 from 1/2 the numerator is 1 why 0??

First application of lhopital gives
(1/(x+1) + A + 2Bx) / 2x
and taking the limit as x->0 gives (1+A)/0. We know the function converges to 1/2, but as the denominator of our expression is 0, we cant get this to match. To justify taking lhopital again, the result must be indeterminant, so 0/0. Therefore A=-1.

Second application of lhopital gives
(-1/(x+1)^2 + 2B) / 2
and taking the limit as x->0 gives (-1+2B)/2. This equals 1/2 when B=1.

There is no setting 2x=2. Here you end up differentiating the numerator and denominator twice, as the first time gives 0/0 which is indeterminant, just like the original function. If youre unsure about any of the above, highlight which parts. But Id play around with a few simpler rational functions such as just linear and quadratic on the numerator and denominator in order to make sure you understand the process.

by doing the rule twice i can easily find first that b=1 but then i dont know to to find a,
sorry i still dont get why a=-1 i have hilighted some the things i did not get....