Question

How do I compute the limit as x -> 3 for
\[(f(x1) - f(x0))\div (x1 -x0) \] ?

Answer 1

\[(f(x1) - f(x0))\div (x1 -x0) \]

Answer 2

and what are x0 and x1 ?

Answer 3

Sorry, I meant as x1 ->x0. x0 is 3.

Answer 4

put x1-3 = h

Answer 5

x1= h+3
as x1->3, h->0

Answer 6

do u recollect the limit definition of derivatives ?

Answer 7

Yes, wouldn't that be
(f(x1) - f(x0))/(x1 -x0)

Answer 8

lim as x1-> x0?

Answer 9

oh yeah, thats also one of the form
so, your limit is just f'(xo), here, f'(3)

Answer 10

Sorry, I forgot to include the function. F(x) = x^2 -1

Answer 11

I can't just plug in the limits, it gives 0/0.

Answer 12

whats your numerator ?

Answer 13

x1^2-9 ?

Answer 14

(f(x1) -f(x0))/(x1-x0) ---> (f(x1) - 8)/(x1 -3) As x1 goes to 3, it becomes 0 on both.

Answer 15

x1^2-3^2= (x1-3)(x1+3)
and x1-3 cancels from both numerator and denom.

wouldn't the numerator be x1^2-9 ??
how u getting 8 ?

Answer 16

I plugged in 3 for x0, and took the limit as x1-->x0. I think that's where I made my mistake.

Answer 17

that is correct, i just want to know how u got 8 in the numeraor..

Answer 18

since it is f(x1) - f(x0), I plugged 3 in for x0. So then I calculated f(3), which is 3^2 -1 which is 8.

Answer 19

what about f(x1) = .... ?

Answer 20

for that, I figured that since x1 -->x0, it would have the same result.

Answer 21

nopes, to get f(x1) from f(x) just put x=x1
so, f(x1) = x1^2-1
got this ?

Answer 22

Yep! I think I can take it from here. I actually have to go.

Answer 23

ok, good luck! :)

Answer 24

Thanks!