- #1
schattenjaeger
- 178
- 0
Limits:( If it were things like derivatives and integrals I'd be good to go but nooooo, stupid limits
lim(x-->0) [(1+ax)^(1/3)-1]/x=1
find a such that this is true. Guessing on my calculator tells me its 3 but I don't know how to do it analytically. My first guess is clear the radical, but I don't know how to clear a cube root without making it worse, if you can
lim(x-->2)[sqrt(6-x)-2]/[sqrt(3-x)-1]=?
again, calculator tells me 1/2, but I can't. Trying to clear either numerator or denominator of its radicals gets either the top or bottom to be 2-x and the other side to be a bigger mess which I can decipher.
and finally, pick two functions f(x) and g(x) such that lim(x-->0)[f(x)+g(x)] exists, but f(x) and g(x)'s limits as x-->0 DON'T exist
I really don't think that's possible and wouldn't know where to start finding those functions if it weren't
lim(x-->0) [(1+ax)^(1/3)-1]/x=1
find a such that this is true. Guessing on my calculator tells me its 3 but I don't know how to do it analytically. My first guess is clear the radical, but I don't know how to clear a cube root without making it worse, if you can
lim(x-->2)[sqrt(6-x)-2]/[sqrt(3-x)-1]=?
again, calculator tells me 1/2, but I can't. Trying to clear either numerator or denominator of its radicals gets either the top or bottom to be 2-x and the other side to be a bigger mess which I can decipher.
and finally, pick two functions f(x) and g(x) such that lim(x-->0)[f(x)+g(x)] exists, but f(x) and g(x)'s limits as x-->0 DON'T exist
I really don't think that's possible and wouldn't know where to start finding those functions if it weren't