# How do you find a power series converging to #f(x)=sinx/x# and determine the radius of convergence?

with radius of convergence

and determine the radius of convergence using the ratio test:

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To find a power series converging to ( f(x) = \frac{\sin x}{x} ) and determine the radius of convergence, you can use Maclaurin series expansion. The Maclaurin series for ( \frac{\sin x}{x} ) is given by:

[ \frac{\sin x}{x} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n + 1)!} ]

To determine the radius of convergence, use the ratio test:

[ R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right| ]

where ( a_{n} ) is the nth term of the series.

In this case, ( a_{n} = \frac{(-1)^n x^{2n}}{(2n + 1)!} ).

Apply the ratio test and find the limit as ( n ) approaches infinity:

[ R = \lim_{n \to \infty} \left| \frac{(-1)^n x^{2n}}{(2n + 1)!} \cdot \frac{(2n - 1)!}{(-1)^{n+1} x^{2(n+1)}} \right| ]

[ R = \lim_{n \to \infty} \left| \frac{x^2}{(2n + 1)(2n + 2)} \right| ]

[ R = |x| \lim_{n \to \infty} \frac{1}{\sqrt{n(n + 1)}} ]

[ R = |x| \cdot 0 ]

[ R = 0 ]

Therefore, the radius of convergence is ( R = 0 ), which means the series converges only at ( x = 0 ).

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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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