no l'hopital rule hard limit

[spice aurora]

Any ideas??

[peak finch]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[drifting jasper]

Try expanding √(1 - x^2) and cos(x) up to their x^4 terms first.

[spice aurora]

What do you mean

[wooden scarab]

Wait a seocnd

im gonna help u

first of all, which indeterminate form?

[spice aurora]

0/0

[wooden scarab]

I'd try to use some algebra to get some special limits (like sinx/x(

idk the right terms

[gusty veldt]

id use this

[wooden scarab]

ah yes

this is what i meant

[drifting jasper]

Recall the following:
(1 + x)^n = 1 + nx + n(n - 1)x^2/2 + o(x^2)
cos(x) = 1 - x^2/2 + x^4/24 + o(x^4)

[spice aurora]

I ain't never seen those formulas in my life

But aight

[wooden scarab]

wait

[wooden scarab]

i almost solved it, but you need to try it too

[wooden scarab]

i get -1

AAAAAAAAAA

[drifting jasper]

Hm, let me try.

[wooden scarab]

lord

try to catch the error here

[drifting jasper]

For x -> 0 we have:
√(1 - x^2) = 1 - x^2/2 - x^4/8 + o(x^4)
cos(x) = 1 - x^2/2 + x^4/24 + o(x^4)
√(1 - x^2) = -x^4/6 + o(x^4)
So:
√(1 - x^2)/cos(x) = 1 + (√(1 - x^2) - cos(x))/cos(x) = 1 + (-x^4/6 + o(x^4))/cos(x), x -> 0
For the cos(x) in the denominator, cos(x) = 1 + o(1) is enough, as taking more terms will produce terms higher than x^4, which won't matter. So:
√(1 - x^2)/cos(x) = 1 + (-x^4/6 + o(x^4))/cos(x) = 1 - (x^4/6 + o(x^4))/(1 + o(1)) = 1 - x^4/6 + o(x^4), x -> 0
Thus:
(1/x^4)ln(√(1 - x^2)/cos(x)) = (1/x^4)ln(1 - x^4/6 + o(x^4)) = (1/x^4)(-x^4/6 + o(x^4)) = -1/6 + o(1) -> -1/6, x -> 0

[drifting jasper]

You forgot to distribute 1/x^4 to ln(cos(x)).

[spice aurora]

What's o

[drifting jasper]

f(x) = o(g(x)), x -> x0 means f(x)/g(x) -> 0 for x -> x0.

[spice aurora]

I don't get it

Maybe I haven't studied it yet

[drifting jasper]

Well, maybe. It's one of the Landau symbols, they are used very often.

[spice aurora]

Is that uni

[drifting jasper]

Yes.

[spice aurora]

I'm in high-school

And that's a me problem

Thanks for your time and knowledge good sir

[wooden scarab]

Lemme try to solve it again

dw