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Let``` ` `f(x)=x^2-2x+8`, what is `lim_(h->0) (f(x+h)-f(x))/h`

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good2beme | Student, Undergraduate | Honors

Posted January 23, 2013 at 3:21 AM via web

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Let``` ` `f(x)=x^2-2x+8`, what is `lim_(h->0) (f(x+h)-f(x))/h`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted January 23, 2013 at 3:30 AM (Answer #1)

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For any function f(x), the limit `lim_(h->0)(f(x+h)-f(x))/h` is the derivative f'(x).

As f(x) = x^2 - 2x + 8, f'(x) = 2x - 2

For the function `f(x) = x^2 - 2x + 8` , `lim_(h->0)(f(x+h)-f(x))/h = 2x - 2`

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Wilson2014 | eNotes Employee

Posted February 20, 2013 at 11:43 PM (Answer #2)

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The above response is correct. `lim_(h->0) (f(x+h)-f(x))/(h)` is the derivative f'(x), but if the function f(x) needs to be evaluated using the limit equation, this is the method to approach the problem.

The limit function tells us that we first need to substitute (x+h) for "x" in f(x), and then subtract it with the actual f(x). Therefore:

`lim_(h->0) (f(x+h)-f(x))/(h)` ` ``= lim_(h->0) ((x+h)^2-2(x+h)+8-(x^2-2x+8))/h`

 

`=> lim_(h->0) (x^2+2xh+h^2-2x-2h+8-x^2+2x-8)/h`

 

After simplifying the fraction, we can take the limit by substituting 0 for h.

`=> lim_(h->0) (2x+h-2)`

`=> 2x+0-2 = 2x-2`

 

Therefore, the answer is 2x-2.

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