Please post or pm if you can help, thank you.

Please post your question.

derive the differentiate divided by the divident modulus p

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hope this helps

Here's one of em

To solve this limit, you need to recall a neat property:
A^3 - B^3 = (A-B)(A^(2)+AB+B^2) and note that x-1 = (x^(1/3)^3 - (1)^3 .

Using this property, we simplify the limit expression, rewriting the numerator as (x^(1/3) - 1) * [(x^(1/3))^2 + (x^(1/3))*1 + (1)^2].
After some work, the expression will ultimately simplify to:

lim [1/(1+x^(1/3)+x^(2/3))], which is no longer in an indeterminate form and is an easy limit to evaluate.
x--> 1


If you want to offer FG, you can send it to him since I clearly have no use for it:
http://forums.d2jsp.org/user.php?i=863241

Solving that limit would make it equal to 1/3, no?
However the answer is 2/3.

e: nevermind, I solved it with l'hospital

Perhaps you can help me with another one.


This post was edited by tonybarb on Sep 25 2016 05:00pm

a = 3, b =6
or basically any number a and b such that sqrt(a+b) = 3 <=> a+b = 9

sqrt [3(1)+6] = sqrt[9] = 3

3-2 / 1 = 1

I got this aswell, would you say this is sufficient work shown?
Impossible to solve 2 variables with only 1 equation, but the two variables added must equal 9. Is there a special notation for this?

I think it would be sufficient to write out an explanation like something you have there.
Just make sure to include the fact that ax+b >= 0.