The average was 79.86%, quite good, although I have to say I was very generous and gave the point when they had the idea even with some parts of the reasoning being wrong on the presentation.




Remarks : 
*** GENERAL REMARK : Lots of reasoning are either very messy and hardly make any sense, or do not contain any word and I have to try and give a sense to their computations myself. In any case I recommend students to try to work together and explain their solutions to each other, it will give them a hint if their explanation is clear enough. They must remember that they don't write to themselves, they write to someone else, who is not in their head.

*** Exercise 1 : lots of (a-b)(a+b)>=0 so (a-b)>=0, you divide by (a+b) so you need to acknowledge you might have a=b=0 and treat it separately


*** Exercise 1 : It is tempting to use the square root on the inequality, a few people did, but using that the square root is increasing is basically just using that the square function is increasing so they are using the result they want to prove

*** Exercise 1 and 2 : lots didn't prove both directions required


*** Too many people try to square inequalities with absolute value, it should come in last resort, it overcomplicates things and leads often to mistakes


*** GENERAL MISTAKE about epsilons and deltas : lots did try to modify what they couldn't, they need to understand that if we want to prove the existence of a limit, we cannot touch at epsilon, we have to find a delta, but if we know the limit exists, we can take any epsilon, and there you cannot touch the delta.

*** some a<b and a<c then b<c (for |f(x)-L|<epsilon so |f(x)|+|L|< epsilon )


*** With the latter mistake and a few others, people had values of delta that could be negative values, or values where delta would always be greater than 1 (which implies f constant on (x-1,x+1)).

It could even be an exercise : if f is not constant around a, i.e. for all delta>0 there is x in (a-delta,a+delta) such that f(x) is not equal to f(a) and if f is continuous at a then there is for all epsilon there is delta(epsilon) such that .... (definition of limit) then delta(epsilon) goes to 0 when epsilon goes to zero.

*** Lots of people didn't try the last exercise. If students have an intuition, then it is better to write it and they get half the points if the intuition is good