# Find the Taylor series expansion formula of #f(x)=\lnx# at #a=3#?

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#f'=1/x=x^-1#

#f''=-1/x^2=(-1)x^-2#

#f'''=2/x^3=(-1)(-2)x^-3#

#f^4=(-1)(-2)(-3)x^-4...#

**must** find #C_n# before finding #f(x)# expansion

**must** find

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To find the Taylor series expansion formula of f(x) = ln(x) at a = 3, we first need to find the derivatives of ln(x) and evaluate them at x = 3.

The derivatives of ln(x) are: f'(x) = 1/x f''(x) = -1/x^2 f'''(x) = 2/x^3 f''''(x) = -6/x^4

Now, let's evaluate these derivatives at x = 3: f'(3) = 1/3 f''(3) = -1/9 f'''(3) = 2/27 f''''(3) = -6/81 = -2/27

Now, let's write the Taylor series expansion formula using these derivatives: f(x) ≈ f(3) + f'(3)(x - 3) + f''(3)(x - 3)^2/2! + f'''(3)(x - 3)^3/3! + f''''(3)(x - 3)^4/4!

Substitute the values we found: f(x) ≈ ln(3) + (1/3)(x - 3) - (1/9)(x - 3)^2/2 + (2/27)(x - 3)^3/6 - (2/27)(x - 3)^4/24

This is the Taylor series expansion formula of f(x) = ln(x) at a = 3.

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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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