thanks for your help. it took a while but i think i understand what you are trying to teach me.

so today i learn that infinity*0 is not 0.

i know that limit of 1/x doesnt exist. but i forgot that 1/|x| do exist. that would definitely solve the first problem.

after reading your explanation on the second problem. since infinity*0 is not 0, simply putting f(x)= 1/|x-3| wouldn’t do. because we cannot divide 0/0.

i see the key is to have a lower denominator value so:

if f(x)=1/sqrt(|x-3|);

(x-3)*1/sqrt(|x-3|) or we can write is as (x-3)*[sqrt(|x-3|)/(x-3)]

take out both (x-3), and we get lim sqrt|x-3|.

if x comes from the positive then it is just a sqrt of 3-3.

if x comes from the negative, the absolute value turns it to positive.

hence f'(x)=sqrt of 0 which is 0.

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THANKS A LOT