# How can you find the taylor expansion of #f(x) =sinx# about x=0?

See the explanation.

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To find the Taylor expansion of ( f(x) = \sin(x) ) about ( x = 0 ), we can use the formula for the Taylor series expansion:

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

For the function ( f(x) = \sin(x) ), the derivatives at ( x = 0 ) are as follows:

- ( f(0) = \sin(0) = 0 )
- ( f'(0) = \cos(0) = 1 )
- ( f''(0) = -\sin(0) = 0 )
- ( f'''(0) = -\cos(0) = -1 )

Substitute these values into the Taylor series expansion formula:

[ \sin(x) = 0 + 1 \cdot x + 0 \cdot \frac{x^2}{2!} - 1 \cdot \frac{x^3}{3!} + \dotsb ]

Simplify the terms:

[ \sin(x) = x - \frac{x^3}{3!} + \dotsb ]

Thus, the Taylor expansion of ( f(x) = \sin(x) ) about ( x = 0 ) is:

[ \sin(x) = x - \frac{x^3}{6} + \dotsb ]

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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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