Answer:
Step-by-step explanation:
WE are given that
. Then, to now the first for terms, we must replace n by 1,2,3,4 respectively. Then




Note that as n increase,
gets closer to 0. So, the limit of this sequence is 0.
Answer:
Step-by-step explanation:
I'm assuming you meant to type in
because you can only have removable discontinuities where there is a rational (fraction) function. Begin by factoring both the numerator and denominator to
and cancelling out like terms would have us eliminating the (x + 3). That is where there is a removable discontinuity. It leaves a hole. The other discontinuity, (x + 1) doesn't cancel out so it is a non-removable discontuinity, which is a vertical asymptote.
The removable discontinuity is at -3. There is no y value at x = -3 (remember there's only a hole here), because -3 causes the denominator to go to 0 and we all know that having a 0 in the denominator of a fraction is a big no-no!!!
Xy - x - y + 1
explanation: distribute (x-1)(y-1) —> x(y-1) - 1 (y-1) —> xy - x - 1(y-1) —> xy - x - y + 1
That is coordinates, when I got that in school I had to do up a table abt it and do a graph