# Let f(x) = 1/x and find the equation of the tangent line to f(x) at a "nice" point near 0.203, how do you use this to approximate 1/0.203?

Choosing:

the tangent line is:

and in fact:

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To find the equation of the tangent line to ( f(x) = \frac{1}{x} ) at a "nice" point near ( x = 0.203 ), we first find the derivative of ( f(x) ), denoted as ( f'(x) ). Then, we evaluate ( f'(x) ) at ( x = 0.203 ) to find the slope of the tangent line.

The derivative of ( f(x) = \frac{1}{x} ) is ( f'(x) = -\frac{1}{x^2} ).

Evaluating ( f'(x) ) at ( x = 0.203 ): [ f'(0.203) = -\frac{1}{(0.203)^2} \approx -24.632 ]

This gives us the slope of the tangent line. To find the equation of the tangent line, we use the point-slope form:

[ y - y_1 = m(x - x_1) ]

Substituting ( x_1 = 0.203 ), ( y_1 = f(0.203) ), and ( m = -24.632 ), we get:

[ y - \frac{1}{0.203} = -24.632(x - 0.203) ]

Solving this equation will give us the equation of the tangent line.

Once we have the equation of the tangent line, we can use it to approximate ( \frac{1}{0.203} ) by plugging in ( x = 0.203 ) into the equation and solving for ( y ).

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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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