求lim(n->∞) [n-√ (n^2-1)]的值 謝謝~?

lim(n->∞) [n-√(n^2-1)]
=lim(n->∞) [n-√(n^2-1)][n+√(n^2-1)]/[n+√(n^2-1)]
=lim(n->∞) (n^2 -n^2 -1)/[n+√ (n^2-1)]
=lim(n->∞) 1/[n+√ (n^2-1)]
=0
因為分母的數字是非常之大，所以接近零

n-sqrt(n^2-1)
=n-n*sqrt(1-1/n^2)
=n(1-sqrt(1-1/n^2)
=n(1-[1-(1/2)(1/n^2)^2-(1/8)(1/n^2)^4-(1/16)(1/n^2)^6-(1/128)(1/n^2)^8....])
=n(1-[1-0-0-0....]) n->inf
=0

Sol
lim(n->∞)_[n－√(n^2－1)]
=lim(n->∞)_[n－√(n^2－1)]*[n+√(n^2－1)]/[n+√(n^2－1)]
=lim(n->∞)_[n^2－(n^2－1)]/[n+√(n^2－1)]
=lim(n->∞)_1/[n+√(n^2－1)]
=0