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



Asked on

2 Answers | Add Yours

justaguide's profile pic

Posted on (Answer #1)

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`

Wilson2014's profile pic

Posted on (Answer #2)

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.

We’ve answered 287,672 questions. We can answer yours, too.

Ask a question