# How do you find the taylor series series at x=0 of #f(x) = 1/(1-2x)#?

Rather than differentiating, just write out a power series that when multiplied by

Then we find:

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To find the Taylor series of ( f(x) = \frac{1}{1 - 2x} ) centered at ( x = 0 ), we can use the formula for the Taylor series expansion, which is given by:

[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \cdot x^n ]

First, let's find the derivatives of ( f(x) ):

[ f'(x) = \frac{d}{dx}\left(\frac{1}{1 - 2x}\right) = \frac{2}{(1 - 2x)^2} ] [ f''(x) = \frac{d^2}{dx^2}\left(\frac{1}{1 - 2x}\right) = \frac{d}{dx}\left(\frac{2}{(1 - 2x)^2}\right) = \frac{8}{(1 - 2x)^3} ] [ f'''(x) = \frac{d^3}{dx^3}\left(\frac{1}{1 - 2x}\right) = \frac{d}{dx}\left(\frac{8}{(1 - 2x)^3}\right) = \frac{48}{(1 - 2x)^4} ]

From the pattern, we can see that the nth derivative of ( f(x) ) evaluated at ( x = 0 ) is ( 2^n ).

Now, let's plug these derivatives into the formula for the Taylor series:

[ f(0) = 1 ] [ f'(0) = 2 ] [ f''(0) = 8 ] [ f'''(0) = 48 ]

Thus, the Taylor series expansion of ( f(x) = \frac{1}{1 - 2x} ) centered at ( x = 0 ) is:

[ f(x) = 1 + 2x + 8x^2 + 48x^3 + \ldots ]

Therefore, the Taylor series at ( x = 0 ) of ( f(x) = \frac{1}{1 - 2x} ) is ( \sum_{n=0}^{\infty} 2^n \cdot x^n ).

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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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