What is cos(x)/x?

lim x->0 of this function can be found by using L'Hopital's Rule. This is where you split up the limit of this function into both the numerator and denominator of the desired function. So you would have (lim x->0 of cos^3(x))/(lim x->0 of 2x) then you take derivatives of both the numerator and denominator separately.

You would get (lim x->0 of - 3 * sin(x) * cos^2(x))/(lim x->0 of 2) which would be 0/2 or just 0.

for your problem that you needed to have checked, your algebra is a little off, 2(1-cos(3x)/3x)=> 2/3(3-cos(3x)/x). Your limit as x goes to 0 of cos(3x)/x is still zero, so you would be left with 2/3(3-0)=2.

As x approaches 0, cos(x) or for that matter cos^3(x) approaches 1.So your fraction approaches 1/0 and therefore there is no limit. (it's infinity from the right and negative infinity from the left.)

Same for the other except the signs change.

I do not know about the first one. But the work on the bottom is wrong because you are not allowed to put a zero on the bottom of a fraction. You need to munipulate it som how. Read the chapter in your textbook for the different ways i only remember multiplying it by an irregular form of one and that is no fun.