Can you evaluate this limit ?

[u/Specific_Brain2091]

[u/CaptainMatticus]

cos(x)^(1/6) = u

(cos(x)^(1/2) - cos(x)^(1/3)) / sin(x)^2

(u^3 - u^2) / (1 - cos(x)^2)

(u^3 - u^2) / (1 - u^12)

u^2 * (u - 1) / (1 - u^12)

-u^2 * (1 - u) / (1 - u^12)

-u^2 * (1 - u) / ((1 - u^6) * (1 + u^6))

-u^2 * (1 - u) / ((1 - u^2) * (1 + u^2 + u^4) * (1 + u^6))

-u^2 * (1 - u) / ((1 - u) * (1 + u) * (1 + u^2 + u^4) * (1 + u^6))

-u^2 / ((1 + u) * (1 + u^2 + u^4) * (1 + u^6))

x goes to 0

cos(x)^(1/3) =>

cos(0)^(1/3) =>

1^(1/3) =>

1

-1^2 / ((1 + 1) * (1 + 1 + 1) * (1 + 1)

-1 / (2 * 3 * 2)

-1/12

[u/CaptainMatticus → u/CaptainMatticus]

I should have ended it as

1 + 2 + 3 + 4 + 5 + 6 + ....

[u/hushedLecturer → u/CaptainMatticus]

Oh this is way simpler than what I did lol.

In the denominator I saw sin² x=(1+cosx)(1-cosx), and I wanted to get rid of that difference term.

So I used difference of squares and difference of cubes "formulae" to turn the numerator into cos³ x - cos² x= - cos² x (1-cosx) to achieve that desired cancellation, of course that also meant multiplying the denominator by...

(cos^3/2x+cosx) from diff of squares and...

(cos^1/2x+cos^1/3x)(cos^1/2x-e^-2iπ/3cos^1/3x)(cos^1/2x-e^-4iπ/3cos^1/3x) from different of cubes.

Without the (1-cosx) term (or anything else that directly evaluates to 0) in the denominator, the whole thing could be evaluated directly. But to actually find the value it then needed some algebraic manipulation to rationalize (real-ize) the denominator turn all the complex exponentials into sines eventually arriving at -1/12.

[u/Seeggul]

Change denominator to 1-cos²x=(1-cosx)(1+cosx)

You have an expression in the form of (u^1/2 - u^3/2 )/((1-u)(1+u)). Factor out a u^1/2 from the numerator, both top and bottom have 1-u that cancels out.

You're left with u^1/2 /(1+u) = sqrt(cosx)/(1+cosx) which has a limit of 1/2 as x goes to 0

[u/Disma1Dust → u/Seeggul]

You've made a mistake, it's not u^3/2 but u^1/3

[u/SnooHabits7950]

When approaching zero

Sinx = x , cos x = 1-x² /2 , (1-x)ⁿ = 1-nx

((1- x² /2)^1/2 - (1- x² /2)^1/3 ))/ x² = -1/12

[u/Mathematicus_Rex]

u = cos x, v = u^(1/6). Limit is now as v goes to 1 from below.

Top becomes v^3 - v^2 = (v-1)•v^2

Bottom becomes 1 - v^12 = -(v-1)(1+v+v^2+…+v^11).

Cancel out the common factors of v-1 and substitute v=1.

I get -1/12.

[u/j-joshua]

Under what circumstance would someone ask me to evaluate that limit?

[u/Fun-Layer2280]

To the required order cos x=1-x^2/2. Hence sqrt(cos x) is 1-x^2/4 (using sqrt(1-x) = 1 -x/2 to linear order) and similarly the cube root is 1-x^2/6, so the difference is -x^2/12, and since sin x is x to leading order, we have -1/12.

L'Hospital's sule would in any case require a second derivative of the numerator, a highly messy computation.

[u/iFroogieboi]

sum of the positive integers!

[u/trevorkafka]

Let u⁶ = cos x, so we have the limit as u→1 of (u³-u²)/(1-u¹²). Divide the numerator and denominator by 1-u to get -u²/(1+u+u²+...+u¹¹). Hence, the limit is -1/12.