The lim h->0 abs(x+h)-abs(x)/h at x=3 is?

- anonymous

The lim h->0 abs(x+h)-abs(x)/h at x=3 is?

- schrodinger

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- anonymous

can i see a pic of you, i will answer your question

- anonymous

uhm a pic of me?

- myininaya

1

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## More answers

- anonymous

yeah

- anonymous

why?

- anonymous

it helps me do math

- anonymous

keeps my mind on the prize,

- myininaya

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

- anonymous

so i don't put 0 in for h?

- myininaya

h/h=1

- anonymous

the derivative is not defined for absolute value at x = 0

- anonymous

oh but near x = 3, thats fine

- anonymous

right the answer is 1

- anonymous

By the definition of a derivative, this is derivative of lxl at x=3. Don't you agree myininaya?

- anonymous

look at the graph of y = | x|

- anonymous

its a V, it has two branches , the negative branch has slope -1, the positive branch has +1 slope

- myininaya

bur since there is a sharp turn at x=0 then f' dne at x=0

- myininaya

so for all x>0 f'(x)=1
for all x<0 f'(x)=-1
f'(0) DNE

- anonymous

Yeah, I agree.

- myininaya

yeah we agree on something lol

- anonymous

:)

- anonymous

anwar, i proved the lim problem

- anonymous

You did?

- anonymous

its a direct result of using the definition of a limit

- anonymous

yeah

- anonymous

How?

- anonymous

You did it or someone did it and you claimed it's you, as you did with mine? :P

- anonymous

well someone said to state the definition of a limit

- anonymous

and then i looked at it, and voila

- anonymous

I see. Btw, I was just kidding.

- anonymous

oh i see, dark humor

- anonymous

LETS do the proof

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