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can i see a pic of you, i will answer your question
uhm a pic of me?
it helps me do math
keeps my mind on the prize,
if x>0 then |x+h|=x+h and |x|=x if x<0 then |x+h|=-(x+h) and |x|=-x but we have the first version so the answer is 1
so i don't put 0 in for h?
the derivative is not defined for absolute value at x = 0
oh but near x = 3, thats fine
right the answer is 1
By the definition of a derivative, this is derivative of lxl at x=3. Don't you agree myininaya?
look at the graph of y = | x|
its a V, it has two branches , the negative branch has slope -1, the positive branch has +1 slope
bur since there is a sharp turn at x=0 then f' dne at x=0
so for all x>0 f'(x)=1 for all x<0 f'(x)=-1 f'(0) DNE
Yeah, I agree.
yeah we agree on something lol
anwar, i proved the lim problem
its a direct result of using the definition of a limit
You did it or someone did it and you claimed it's you, as you did with mine? :P
well someone said to state the definition of a limit
and then i looked at it, and voila
I see. Btw, I was just kidding.
oh i see, dark humor
LETS do the proof