# How do you find the taylor series series for # (4-x)^(1/2)#?

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To find the Taylor series for ( (4-x)^{1/2} ), follow these steps:

- Determine the function and the point about which you want to expand the Taylor series. In this case, the function is ( (4-x)^{1/2} ) and the point is ( x = 0 ).
- Find the derivatives of the function up to the desired order.
- Evaluate the derivatives at the point of expansion (( x = 0 )).
- Write out the Taylor series using the formula:

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

where ( f(a) ) is the value of the function at the point of expansion and the derivatives are evaluated at that point.

For ( (4-x)^{1/2} ):

- ( f(0) = (4-0)^{1/2} = 2 )
- ( f'(x) = -\frac{1}{2}(4-x)^{-1/2} )
- ( f''(x) = \frac{1}{4}(4-x)^{-3/2} )
- ( f'''(x) = -\frac{3}{8}(4-x)^{-5/2} )

Now evaluate these derivatives at ( x = 0 ):

- ( f'(0) = -\frac{1}{2}(4-0)^{-1/2} = -\frac{1}{4} )
- ( f''(0) = \frac{1}{4}(4-0)^{-3/2} = \frac{1}{8} )
- ( f'''(0) = -\frac{3}{8}(4-0)^{-5/2} = -\frac{3}{32} )

Plug these values into the Taylor series formula:

[ (4-x)^{1/2} = 2 - \frac{1}{4}x + \frac{1}{8}x^2 - \frac{3}{32}x^3 + \cdots ]

This is the Taylor series expansion for ( (4-x)^{1/2} ) centered 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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