# How do you find the Taylor remainder term #R_3(x;0)# for #f(x)=1/(2+x)#?

This exact value of

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To find the Taylor remainder term ( R_3(x;0) ) for the function ( f(x) = \frac{1}{2+x} ) centered at ( x = 0 ), we can use the formula for the remainder term in Taylor's theorem:

[ R_n(x;0) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} ]

where ( c ) is some point between ( 0 ) and ( x ).

First, we find the first four derivatives of ( f(x) ):

[ f(x) = \frac{1}{2+x} ] [ f'(x) = -\frac{1}{(2+x)^2} ] [ f''(x) = \frac{2}{(2+x)^3} ] [ f'''(x) = -\frac{6}{(2+x)^4} ]

Now, we evaluate ( f(x) ) and its derivatives at ( x = 0 ) to find the values of the derivatives:

[ f(0) = \frac{1}{2} ] [ f'(0) = -\frac{1}{4} ] [ f''(0) = \frac{1}{4} ] [ f'''(0) = -\frac{3}{16} ]

Now, substitute these values into the remainder term formula for ( n = 3 ):

[ R_3(x;0) = \frac{f'''(c)}{4!}x^4 ]

To find ( c ), note that it lies between ( 0 ) and ( x ). Since we're centered at ( x = 0 ), ( c ) will be somewhere between ( 0 ) and ( x ).

[ R_3(x;0) = \frac{-\frac{3}{16}}{4!}x^4 ] [ R_3(x;0) = -\frac{3}{16 \times 24}x^4 ] [ R_3(x;0) = -\frac{1}{128}x^4 ]

So, the Taylor remainder term ( R_3(x;0) ) for ( f(x) = \frac{1}{2+x} ) is ( -\frac{1}{128}x^4 ).

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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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