Question

i am going to produce a silly piecewise function, but the point is to show that limit definition of derivative has a problem

let f(x) = {1 if x<0 , 1 if x >= 0}

Answer 1

and we want f ' (0) = 0 , since f(x) = 1 is a constant function

actually the proof works with just a simple constant function f(x) = 1

Answer 2

-.-
is this is ur function ?

Answer 3

yes

Answer 4

now i claim that the derivative exists at x =0,
but we have a problem
f ' (0) = lim [ f( 0 + h )- f(0)] / h , h->0

lets look at the right hand limit
lim h->0+ [ f( 0 + h )- f(0)] / h
=lim h->0+ [f(h) -f(0 )] / h
= lim h->0+ (1 - 1) / h
= lim h->0+ 0 /h

but limit 0/h as h->0+ is indeterminate

Answer 5

that was a typo, it should say 0/h

Answer 6

maybe because in the limit, h does not equal to zero?

Answer 7

so here u have mistake here
http://prntscr.com/5fnmon

Answer 8

oh u already set u made a typo,
then its like this
lim (0/h) = lim (0) =0

Answer 9

"but limit 0/h as h->0+ is indeterminate "
not true , since 0/h=0 so u dont need to define what h is

Answer 10

i am using derivative definition
f ' (x) = lim h->0 [ f( x + h ) - f(x) ] / h

so
f ' (0) = lim h->0 [ f( 0 + h ) - f(0) ] / h

Answer 11

yeah its ovc u used it !
u would end up with
f'(0)=lim 0/h =lim 0 =0

Answer 12

and I am evaluating the right hand derivative

my logic was

lim {h->0+} 0/h
= lim {h->0+} 0* 1/h
= 0 * infinity

Answer 13

ugh ok w.e

Answer 14

lim {h->0+} 0/h = lim {h->0+} 0 = 0

Answer 15

numerator is a fixed value, it is not "tending" to 0, it IS 0.

Answer 16

what happened to your division by h?

Answer 17

lim {h->0+} [ f( 0 + h )- f(0)] / h
=lim {h->0+} [f(h) -f(0 )] / h
= lim {h->0+} (1 - 1) / h
= lim {h->0+} 0 /h
= lim {h->0+} 0 * 1/h

and the latter expression i have a problem with

i agree the numerator is 0.

Answer 18

its only algebraic trick :) has nothing to do with derivative

Answer 19

h goes to 0 , it doesnt equal zero so its 0 over h , and h is a constant

Answer 20

ok according to wolfram it is true that
http://www.wolframalpha.com/input/?i=lim+%28x-%3E0%2B%29+0%2Fx

Answer 21

marki, i think thats it , since h does not equal to zero
as h->0+ , h>0
as h->0- , h < 0

but h is not equal to zero

Answer 22

0/h = 0 , for h =/= 0

I wasn't sure if this is correct

Answer 23

oh , so now u are sure ?

Answer 24

hehe

Answer 25

maybe :)

Answer 26

remember this :-
when u have limits dont get scared of such things :D
limits are used to deal with such situations

Answer 27

ok but if you do this argument, weird things happen

Answer 28

= lim {h->0+} 0 /h
= lim {h->0+} 0 * 1/h
=lim {h->0+} 0 * lim {h->0+} (1/h)
= 0 * infinity

Answer 29

well , i can see these things easily which is correct and which is not ,
so its not a problem for me , i only have a problem how to explain :P

Answer 30

k, got what u are saying :)

Answer 31

then i guess we shouldn't do that?

Answer 32

indeterminate limits are 'ok', we just have to find a way to simplify

Answer 33

yeah , btw im looking at the question again
and i guess its very important thing for students to understand and explain

Answer 34

thanks :)

Answer 35

also the question would have been easier if we just defined f(x) = 1 for all x

Answer 36

actually i just realized i did an invalid move.
it is false that
= lim {h->0+} 0 * 1/h =lim {h->0+} 0 * lim {h->0+} (1/h)

because
lim {x->a} (f(x) * g(x)) = lim {x->a} f(x) * lim {x->a} g(x) , provided lim f(x), lim g(x) exist

in our example lim 1/h as h->0+ does not exist

Answer 37

provided lim f(x), lim g(x) exist as x->a+ *

Answer 38

@FibonacciChick666
I was inspired to post this problem after working on your problem :)

Answer 39

:) I spent three hours and balled my eyes out in anxiety after that problem, I finished the test though...hoping for at least a 50%