sqrt(1+tan x)-sqrt(1+sin x)

lim all divided by x^3
x-->0

Use that Sqrt[1+x] = 1+ 1/2 x + 1/2 (-1/2)/2 x^2 + 1/2(-1/2)(-3/2)/6 x^3 +
O(x^4)

You can thus write the numerator as:

1/2 [tan(x) - sin(x)] - 1/8 [tan^2(x) - sin^2(x)] + 1/16 [tan^3(x) - sin^3(x)] +
O(x^4)

Note that sin(x) and tan(x) are of order x near x = 0 so that the neglected terms are of order x^4 and won't contribute to the limit. The next step is to insert the series expansions of the sin and the tan. The series expansion of the tan can be obtained by division. We only need to work to order x^3, so:

tan(x) = sin(x)/cos(x) = (x - x^3/6)/(1-x^2/2) + O(x^5) =

(x - x^3/6)(1 + x^2/2) + O(x^5) =

x + x^3/3 + O(x^5)

So, you see that in the first square brackets you get an term of order x^3 and you can divide by x^3 to get a finite limit for x --> 0 (I find 1/12). The third square brackets yields a term of order x^5 and thus also doesn't contribute to the limit.

The term in the second square brackets is of order x^4 and thus yields zero when divided by x^3 and x -->0.

To solve this limit, we need to simplify the expression using the given series expansion. Let's start by expanding the numerator using the provided series expansion of the square root function:

sqrt(1+tan x) = 1 + 1/2 * x + 1/2 * (-1/2)/2 * x^2 + 1/2 * (-1/2) * (-3/2)/6 * x^3 + O(x^4)
sqrt(1+sin x) = 1 + 1/2 * x + 1/2 * (-1/2)/2 * x^2 + 1/2 * (-1/2) * (-3/2)/6 * x^3 + O(x^4)

Next, let's simplify the numerator:

1/2 * [tan(x) - sin(x)] - 1/8 * [tan^2(x) - sin^2(x)] + 1/16 * [tan^3(x) - sin^3(x)] + O(x^4)

Now, let's substitute the series expansion of tan(x):

tan(x) = x + x^3/3 + O(x^5)

Plugging this into our numerator, we get:

1/2 * [(x + x^3/3 + O(x^5)) - sin(x)] - 1/8 * [(x + x^3/3 + O(x^5))^2 - sin^2(x)] + 1/16 * [(x + x^3/3 + O(x^5))^3 - sin^3(x)] + O(x^4)

To simplify further, we can use the Taylor expansion of sin(x):

sin(x) = x - x^3/6 + O(x^5)

Continuing with the numerator:

1/2 * [(x + x^3/3 + O(x^5)) - (x - x^3/6 + O(x^5))] - 1/8 * [(x + x^3/3 + O(x^5))^2 - (x - x^3/6 + O(x^5))^2] + 1/16 * [(x + x^3/3 + O(x^5))^3 - (x - x^3/6 + O(x^5))^3] + O(x^4)

Expanding and collecting like terms, we get:

1/2 * 2x^3/3 - 1/8 * 2x^3/3 + O(x^4)
= x^3/3 - x^3/12 + O(x^4)
= 3x^3/12 - x^3/12 + O(x^4)
= 2x^3/12 + O(x^4)
= x^3/6 + O(x^4)

Now, we can simplify the expression further by dividing the numerator by x^3:

(x^3/6 + O(x^4)) / x^3
= (1/6) + O(x)

Finally, we can take the limit as x approaches 0:

lim x-->0 ((1/6) + O(x))
= 1/6

Therefore, the limit of sqrt(1+tan x) - sqrt(1+sin x) / x^3 as x approaches 0 is 1/6.