# How do you write the taylor series about 0 for the function #(1+x)^3#?

The Taylor series computes as

Then:

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To find the Taylor series about (x = 0) for the function ((1 + x)^3), we can use the general formula for the Taylor series expansion:

[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots ]

For this function, (f(x) = (1 + x)^3), and we're expanding around (a = 0). We'll need the derivatives of (f(x)) evaluated at (x = 0).

- (f(0) = (1 + 0)^3 = 1)
- (f'(x) = 3(1 + x)^2), so (f'(0) = 3(1 + 0)^2 = 3)
- (f''(x) = 6(1 + x)), so (f''(0) = 6(1 + 0) = 6)
- (f'''(x) = 6), so (f'''(0) = 6)

Now we plug these values into the Taylor series formula:

[ (1 + x)^3 = 1 + 3x + \frac{6}{2!}x^2 + \frac{6}{3!}x^3 + \cdots ]

Simplify:

[ (1 + x)^3 = 1 + 3x + 3x^2 + x^3 + \cdots ]

This is the Taylor series expansion of ((1 + x)^3) about (x = 0).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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