# How do you find a power series representation for #f(x)= 1/(3-x)# and what is the radius of convergence?

The radius of convergence is -3< x <3

The radius of convergence is therefore -3< x <3

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To find a power series representation for ( f(x) = \frac{1}{3-x} ), we can use the geometric series formula ( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n ) for ( |x| < 1 ). Let's rewrite ( f(x) ) in a form suitable for the geometric series:

[ f(x) = \frac{1}{3-x} = \frac{1}{3(1-\frac{x}{3})} ]

Now, replace ( \frac{x}{3} ) with ( u ), which gives ( x = 3u ):

[ f(x) = \frac{1}{3(1-u)} = \frac{1}{3} \cdot \frac{1}{1-u} ]

This is in the form ( \frac{1}{1-x} ), so we can apply the geometric series formula:

[ \frac{1}{3} \cdot \sum_{n=0}^{\infty} u^n = \sum_{n=0}^{\infty} \frac{u^n}{3} ]

Substitute ( u = \frac{x}{3} ) back in:

[ \sum_{n=0}^{\infty} \frac{x^n}{3^{n+1}} ]

The radius of convergence can be found using the ratio test. Let's apply the ratio test to the series:

[ a_n = \frac{1}{3^{n+1}} ]

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{1}{3^{n+2}} \cdot \frac{3^n}{1} \right| = \frac{1}{3} ]

Since the limit is less than 1, the series converges for all ( x ) within a radius of ( R = 3 ). Thus, the radius of convergence is 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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