# How do you find the taylor series series for #f(x)=lnx# at a=2?

First, we can start with the general definition of the Taylor series expansion, which is:

Now you have the final result to simplify:

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Therefore, we can write the answer as

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To find the Taylor series for ( f(x) = \ln(x) ) at ( a = 2 ), we start by finding the derivatives of ( f(x) = \ln(x) ) and evaluating them at ( x = 2 ) to obtain the coefficients of the Taylor series.

- First derivative of ( f(x) = \ln(x) ): ( f'(x) = \frac{1}{x} )
- Second derivative of ( f(x) = \ln(x) ): ( f''(x) = -\frac{1}{x^2} )
- Third derivative of ( f(x) = \ln(x) ): ( f'''(x) = \frac{2}{x^3} )
- Fourth derivative of ( f(x) = \ln(x) ): ( f''''(x) = -\frac{6}{x^4} )

Now, we evaluate these derivatives at ( x = 2 ):

- ( f'(2) = \frac{1}{2} )
- ( f''(2) = -\frac{1}{4} )
- ( f'''(2) = \frac{1}{4} )
- ( f''''(2) = -\frac{3}{16} )

Since ( f(a) = \ln(2) ) when ( a = 2 ), the first term of the Taylor series is ( \ln(2) ).

Therefore, the Taylor series for ( f(x) = \ln(x) ) at ( a = 2 ) is:

[ \ln(x) = \ln(2) + \frac{1}{2}(x - 2) - \frac{1}{8}(x - 2)^2 + \frac{1}{12}(x - 2)^3 - \frac{1}{32}(x - 2)^4 + \cdots ]

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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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