# How do you find the maclaurin series expansion of #f(x) = x/(1-x)#?

If

You must compute the derivatives, evaluate them in zero, and then apply the expression to understand why this is the case.

That's all there is to it if your main concern is how you had to discover the expansion; if you would like verification of the last point, please ask.

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To find the Maclaurin series expansion of ( f(x) = \frac{x}{1-x} ), we'll start by expressing it as a geometric series.

[ f(x) = x \cdot \frac{1}{1-x} ]

Now, we'll use the geometric series formula:

[ \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n ]

Multiplying by ( x ):

[ x \cdot \frac{1}{1 - x} = x \cdot \sum_{n=0}^{\infty} x^n ]

[ = \sum_{n=0}^{\infty} x^{n+1} ]

This series represents ( f(x) ) in terms of its Maclaurin series expansion. However, we need to make sure that it converges. The radius of convergence ( R ) for this series is 1, which means it converges for ( |x| < 1 ).

So, the Maclaurin series expansion of ( f(x) = \frac{x}{1-x} ) is:

[ \sum_{n=0}^{\infty} x^{n+1} ]

This can also be written as:

[ \sum_{n=1}^{\infty} x^n ]

This series converges for ( |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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