# How do you use the Maclaurin series for # f(x) = (2+x)^5#?

based on the Maclaurin series

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To use the Maclaurin series for ( f(x) = (2+x)^5 ), we need to express ( f(x) ) as a power series centered at ( x = 0 ) (also known as the Maclaurin series). The Maclaurin series representation of a function ( f(x) ) is given by:

[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots ]

For ( f(x) = (2+x)^5 ), let's find the derivatives up to the required order and evaluate them at ( x = 0 ):

- ( f(x) = (2+x)^5 )
- ( f'(x) = 5(2+x)^4 )
- ( f''(x) = 20(2+x)^3 )
- ( f'''(x) = 60(2+x)^2 )
- ( f''''(x) = 120(2+x) )
- ( f'''''(x) = 120 )

Now, evaluate these derivatives at ( x = 0 ):

- ( f(0) = (2+0)^5 = 32 )
- ( f'(0) = 5(2+0)^4 = 80 )
- ( f''(0) = 20(2+0)^3 = 160 )
- ( f'''(0) = 60(2+0)^2 = 240 )
- ( f''''(0) = 120(2+0) = 240 )
- ( f'''''(0) = 120 )

Now, plug these values into the Maclaurin series formula:

[ f(x) = 32 + 80x + \frac{160}{2!}x^2 + \frac{240}{3!}x^3 + \frac{240}{4!}x^4 + \frac{120}{5!}x^5 ]

Simplify this expression to obtain the Maclaurin series for ( f(x) = (2+x)^5 ).

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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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