MQ43 --- Limit

Method 1

Consider y = n1/n

ln y = ln n/nlim (n->∞) ln n/n= lim (n->∞) 1/n (L' Hospital rule)

= 0

So, lim (n->∞) n1/n = e^0 = 1

Method 2

Consider sequence n, √n, 1, 1, ..., 1 (1 being repeated n-2 times).

apply AM ≥ GM inequality

(2√n + n-2)/n ≥ n1/n ≥ 1

i.e. 1+ 2/√n - 2/n ≥ n1/n ≥ 1

Thus, by sandwich theorem n1/n -> 1 as LHS -> 1 and RHS -> 1.

圖片參考：http://imgcld.yimg.com/8/n/HA00100646/o/701209070058413873365520.jpg

Hope I can help you.