# How do you find the derivative of # 1/(x^2-1)# using the limit definition?

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To find the derivative of ( \frac{1}{x^2-1} ) using the limit definition, we start with the limit as ( h ) approaches 0 of the difference quotient:

[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} ]

Substitute ( f(x) = \frac{1}{x^2-1} ) into the difference quotient:

[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2-1} - \frac{1}{x^2-1}}{h} ]

Combine the fractions under the same denominator:

[ f'(x) = \lim_{h \to 0} \frac{(x^2-1) - ((x+h)^2-1)}{h((x+h)^2-1)(x^2-1)} ]

Expand and simplify the numerator:

[ f'(x) = \lim_{h \to 0} \frac{x^2-1 - (x^2 + 2xh + h^2 - 1)}{h((x+h)^2-1)(x^2-1)} ] [ f'(x) = \lim_{h \to 0} \frac{-2xh - h^2}{h((x+h)^2-1)(x^2-1)} ]

Cancel out the common factor of ( h ) in the numerator and denominator:

[ f'(x) = \lim_{h \to 0} \frac{-2x - h}{((x+h)^2-1)(x^2-1)} ]

Now, take the limit as ( h ) approaches 0:

[ f'(x) = \frac{-2x - 0}{((x+0)^2-1)(x^2-1)} ] [ f'(x) = \frac{-2x}{(x^2-1)^2} ]

Therefore, the derivative of ( \frac{1}{x^2-1} ) using the limit definition is ( \frac{-2x}{(x^2-1)^2} ).

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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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