Limits

[hoary inlet]

bacc (unhelpful)

[hasty hill]

i got a to be 1

"a"

[covert marlin]

hmm okay why

also I would split the term

[hasty hill]

i did l'hopital but im stuck with b

[hoary inlet]

bacc (unhelpful)

[hasty hill]

yea

now do it again and you get a*cos(x)/(cos(2x)

[covert marlin]

no

[hasty hill]

?

ok but why did you split the fraction

[covert marlin]

I thought it could lead to something

but how would you apply L'Hopital again?

Unless b = 0 then yes

[covert marlin]

ok good then lol

then there is your answer

if b = 0, then you can apply l'hopital

which gives you then a = 1

[covert marlin]

I think if b is not 0 you would be stuck here

you would basically have the form 1/0 diverging basically in the origin

So I don't think there are infinite solutions

[hasty hill]

but when you check in the original function and do the limit it does not give 1

[covert marlin]

you sure?

[unreal coyote]

i would use 1-cos(x)=2sin(x/2)^2. Then the rest is easy

or if you know about equivalences and Taylor series cos(x)=1-x^2/2+O(x^4).

[hasty hill]

not yet

[covert marlin]

I see

your a must be 1/2

2a = 1 basically

[hasty hill]

well, but if x tends to 0, and both “a” and “b” are accompanied by x both values do not matter since they will cancel each other out and I will still get an indeterminacy.

[covert marlin]

After this you'd get [ \lim_{x \to 0} \frac{2a+\cos(x)}{2\cos(2x)} = \frac{2a+1}{2}=1 ]

[hoary inlet]

bacc (unhelpful)

[covert marlin]

assuming b = 0 ofc

[hasty hill]

or any number actually

[covert marlin]

nope

then you cannot apply l'hopital

[hasty hill]

still get indet

[covert marlin]

you wouldn't have 0/0 form

[hasty hill]

yeah thats right

i see it

the issue

that has me stuck

is that no matter which value you give to a or b

still get indet

so theres no value that make that function equal to one in x =0