Okay, i posted this question yesterday, however, I did not really understand the answer I received. If your the one who answered my question, could you please elaborate.

If not, could you try to answer this tough, for me, question. Thanks a lot.

lim
x-->0

Square root (1 + tan x) - Square root (1 + sin x) all over x^3.

Limits involving analytical functions are most easily calculated using series expansion techniques. In this case we want to calulate a limit for x--> 0, so the natural thing to do is to examine how the functions that are involved in this limit behave in the vicinity of x = 0.

In this problem you need to use that:

sin(x) = x - x^3/6 + term of order x^5

cos(x) = 1 - x^2/2 + term of order x^4

You find tan(x) by dividing sin(x) by cos(x).

Both tan(x) and sin(x) tend to zero for x -->0 so you need to know how the square root function behaves near 1. For y close to zero we have:

sqrt[1 + y] = 1 + 1/2 y + (1/2)(-1/2)/2 y^2 + (1/2)(-1/2)(-3/2)/6 y^3+ term of order y^4

You just insert the series expansion for the tan and sin in y and subrtact the two terms...

It is possible to calculate limits without using series expansions. But that's more tedious because ultimately you must make use of the way the fiunctions behave near zero. If you don't make use of the series expansions then the information contained in them must be derived in some form during that derivation.

Let me give some other examples:

Lim x-->0 sin(x)/x = 1

If you use that sin(x) = x + term of order x^3, then you immediately see that:

sin(x)/x = 1 + term of order x^2

Take limit x-->0 of both sides and you find the desired result.

Lim x-->0 (sin(x)-x)/x^3 = -1/6

Use that:

sin(x) - x = - x^3/6 + term of order x^5

Divide both sides by x^3:

(sin(x) - x)/x^3 =
- 1/6 + term of order x^2

Take the limit x-->0 to find te desired result.

You can also calculate the latter limit by applying L'Hopital's rule three times! :)

I apologize, but I am not the one who answered your question previously. However, I can try to explain the process of solving the limit you provided.

To start, we need to analyze the behavior of the functions involved in the limit near x = 0. In this case, we have the functions sin(x), tan(x), and the square root function.

We can use series expansion techniques to approximate these functions near x = 0. For example:

1. sin(x) can be approximated as: sin(x) = x - (x^3)/6 + terms of order x^5.
2. cos(x) can be approximated as: cos(x) = 1 - (x^2)/2 + terms of order x^4.

By dividing sin(x) by cos(x), we can determine the approximation for tan(x):

3. tan(x) = sin(x) / cos(x) ≈ (x - (x^3)/6 + terms of order x^5) / (1 - (x^2)/2 + terms of order x^4).

Next, we can approximate the square root function near 1. For values of y close to zero, we have:

4. sqrt(1 + y) ≈ 1 + (1/2)y + ((1/2)(-1/2)/2)y^2 + ((1/2)(-1/2)(-3/2)/6)y^3 + terms of order y^4.

Now, we can substitute the series expansions for sin(x) and tan(x) into the expression:

sqrt(1 + tan(x)) - sqrt(1 + sin(x)) = sqrt(1 + ((x - (x^3)/6 + terms of order x^5) / (1 - (x^2)/2 + terms of order x^4))) - sqrt(1 + (x - (x^3)/6 + terms of order x^5)).

By subtracting these two terms, we can simplify the expression and find the series expansion for it. This can involve some arithmetic calculations and algebraic manipulations.

Alternatively, you can use L'Hopital's Rule to evaluate the limit. L'Hopital's Rule states that if the limit of a function f(x) / g(x) as x approaches a is of the form 0/0 or ∞/∞, then the limit of f(x) / g(x) as x approaches a is equal to the limit of f'(x) / g'(x) as x approaches a, provided that the latter limit exists. By applying L'Hopital's Rule three times, you can calculate the limit (sin(x) - x) / x^3 = -1/6 as x approaches 0.

I hope this explanation helps you understand the process of solving the given limit. If you have any further questions, feel free to ask.