How do you find taylor polynomials of degree n approximating #5/(2-2x)# for x near 0?
Hence:
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To find the Taylor polynomial of degree (n) approximating (5/(2-2x)) for (x) near 0, we can use the formula for the Taylor series expansion of a function (f(x)) centered at (a):
[ P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n ]
In this case, (f(x) = 5/(2-2x)), (a = 0), and we want the Taylor polynomial of degree (n) near (x = 0). The derivatives of (f(x)) are:
[ f'(x) = \frac{10}{(2-2x)^2} ] [ f''(x) = \frac{40}{(2-2x)^3} ] [ f'''(x) = \frac{240}{(2-2x)^4} ] [ \vdots ]
Evaluating these derivatives at (a = 0), we get:
[ f(0) = 5/2 ] [ f'(0) = 5 ] [ f''(0) = 20 ] [ f'''(0) = 120 ] [ \vdots ]
Substitute these values into the Taylor polynomial formula to get:
[ P_n(x) = \frac{5}{2} + 5x + \frac{20}{2!}x^2 + \frac{120}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n ]
Which simplifies to:
[ P_n(x) = \frac{5}{2} + 5x + 10x^2 + \frac{40}{3}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n ]
This is the Taylor polynomial of degree (n) approximating (5/(2-2x)) for (x) near 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.
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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