Calculate lim┬(n→∞)〖(√((n^2+n))〗-n) justify your answer.
all those (unbalanced) parentheses! I suspect a typo. I'll try
lim(n→∞)√(n^2+n)-n
= n(√(1+1/n) - 1)
= (√(1+1/n)-1)/(1/n)
Let u = 1/n and you have
(√(1+u)-1)/u
= (√(1+u)-1)(√(1+u)+1) / (u√(1+u)+1)
= (1+u-1)/(u√(1+u)+1)
= u/(u(√(1+u)+1))
= 1/(√(1+u)+1)
lim as n→∞ = lim as u→0, so
= 1/2