Please, explain this step in solution to limit problem?

Question

Sorry had to repost this, made a big mistake:

I asked this question on answers:
find the limit as n->+infinity:

Lim ((n^3 - 5n^2 + n + 2)^(1/3)-n)

I got the following answer:
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Let's make life simple.
As n --> infinity, the smaller powers of n become negligible
n^3 - 5n^2 + n + 2 --> n^3 - 5n^2 = n^3 * (1 - 5/n)
So the expression tends to
n*[(1-5/n)^1/3] - n
= n* [ 1 - 5/(3n) + O(1/n^2)] - n
= n - 5/3 - n
= -5/3
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which is right, but I have no idea how he went from:
=n*[(1-5/n)^1/3] - n to: = n* [ 1 - 5/(3n) + O(1/n^2)] - n

I really want to be able to understand this, so that I can do it in the future myself without help, but I have no idea what he did.

Answer 1

Looks like Dr. D sprung a binomial series on you. Briefly, it so happens that for any value of r, the function f(x) = (1+x)^r has a Taylor series given by:

[n=0, ∞]∑(r, n) x^n

Where (r, n) is the generalized binomial coefficient, given by:

1/n! * [k=0, n-1]∏(r-k)

What Dr. D did was he expanded the formula (1-5/n+1/n²+2/n³)^(1/3) using the binomial series, so that, if he hadn't neglected the 1/n² and 1/n³ terms, we would have:

(1-(5/n+1/n²+2/n³))^(1/3) ≈ 1 - 1/3 (5/n + 1/n² + 2/n³) + stuff

Where stuff consists only of terms whose denominator is n² or higher, so even after multiplying by n, would still approach 0 as n→∞.

Of course, you realize from looking at this that 1/3 * (1/n² + 2/n³) also is O(1/n²), and thus could be included in "stuff" as well. Dr. D realized this as well, which is why he neglected them from the very beginning.

Taylor series are sometimes the simplest way to solve limit problems, and in some cases, the only way. In this case, though, we actually could have solved it directly, using the identity (x-y)(x²+xy+y²) = x³ - y³. Observe:

[n→∞]lim ∛(n³ - 5n² + n + 2) - n

[n→∞]lim (n³ - 5n² + n + 2 - n³)/((∛(n³ - 5n² + n + 2))² + n∛(n³ - 5n² + n + 2) + n²)

[n→∞]lim (-5n² + n + 2)/((∛(n³ - 5n² + n + 2))² + n∛(n³ - 5n² + n + 2) + n²)

[n→∞]lim (-5 + 1/n + 2/n²)/((1/n ∛(n³ - 5n² + n + 2))² + 1/n ∛(n³ - 5n² + n + 2) + 1)

[n→∞]lim (-5 + 1/n + 2/n²)/((∛(1 - 5/n + 1/n² + 2/n³))² + ∛(1 - 5/n + 1/n² + 2/n³) + 1)

(-5 + 0 + 0)/((∛(1 - 0 + 0 + 0))² + ∛(1 - 0 + 0 + 0) + 1)

-5/((∛1)² + ∛1 + 1)

-5/3

Answer 2

I don't think that -5/3 is correct:

As n approaches infinity this:

(n^3 - 5n^2 + n + 2)^(1/3)

approaches n

because this:

(n^3 - 5n^2 + n + 2)

approaches n^3. All of the lower power terms become negligible.

The entire expression approaches 0 because we are subtracting n from n. If I am reading the formula correctly I'm sure I am right. When you plug large numbers into the equation do you see the result approaching 0?

Answer 3

The formula is (1+t)^alpha = 1 + alpha t + O(t^2)
In your case t = 5/n alpha = 1/3
The formula is obtained from Taylor formula
f(t) = f(0) + f'(0) t + O(t^2)
for the function f(t) = (1+t)^alpha