# Using the limit definition, how do you find the derivative of #F(x)=x^3−7x+5#?

by definition

when expanding out that binomial, Pascal's Triangle is handy to know.

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To find the derivative of ( F(x) = x^3 - 7x + 5 ) using the limit definition, we use the formula for the derivative:

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

First, we substitute ( f(x) = x^3 - 7x + 5 ) into the formula:

[ f'(x) = \lim_{{h \to 0}} \frac{{(x + h)^3 - 7(x + h) + 5 - (x^3 - 7x + 5)}}{h} ]

Next, we expand and simplify the numerator:

[ f'(x) = \lim_{{h \to 0}} \frac{{x^3 + 3x^2h + 3xh^2 + h^3 - 7x - 7h + 5 - x^3 + 7x - 5}}{h} ]

Combine like terms and cancel out the common terms:

[ f'(x) = \lim_{{h \to 0}} \frac{{3x^2h + 3xh^2 + h^3 - 7h}}{h} ]

Now, factor out an ( h ) from the numerator:

[ f'(x) = \lim_{{h \to 0}} \frac{{h(3x^2 + 3xh + h^2 - 7)}}{h} ]

Cancel out the ( h ) in the numerator and denominator:

[ f'(x) = \lim_{{h \to 0}} (3x^2 + 3xh + h^2 - 7) ]

Now, substitute ( h = 0 ) into the expression:

[ f'(x) = 3x^2 - 7 ]

So, the derivative of ( F(x) = x^3 - 7x + 5 ) using the limit definition is ( f'(x) = 3x^2 - 7 ).

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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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