# How do you find the Taylor series for #ln(x)# about the value x=1?

firstly we look at the formula for the Taylor series, which is:

which equals:

To solve:

Where now we can already start to see a pattern forming, so we starting using our formula(2):

and now try try see how we can write this as a series, which we get: (we start will n=1 as our first term is 0)

Which can then simplify to:

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To find the Taylor series for ln(x) about the value x=1, you can start by finding the derivatives of ln(x) and evaluating them at x=1 to obtain the coefficients of the Taylor series. The Taylor series for ln(x) about x=1 is given by:

ln(x) = ln(1) + (x - 1) * ln'(1) / 1! + (x - 1)^2 * ln''(1) / 2! + (x - 1)^3 * ln'''(1) / 3! + ...

To find the coefficients, you'll need to find the derivatives of ln(x) and evaluate them at x=1:

ln'(x) = 1/x ln''(x) = -1/x^2 ln'''(x) = 2/x^3

Evaluate these derivatives at x=1:

ln'(1) = 1 ln''(1) = -1 ln'''(1) = 2

Substitute these values into the Taylor series formula:

ln(x) = 0 + (x - 1) * 1 / 1! + (x - 1)^2 * (-1) / 2! + (x - 1)^3 * 2 / 3! + ...

Simplify:

ln(x) = (x - 1) - (x - 1)^2 / 2 + (x - 1)^3 / 3 - ...

This is the Taylor series for ln(x) about the value x=1.

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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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