推 terry1122:用羅必達下手= = 59.104.225.208 02/19
→ terry1122:我還是不會處理分子耶... 59.104.225.208 02/19
※ 引述《terry1122 (我會一直守候著你)》之銘言：
: l i m {x+[x+(x^1/2)]^1/2}^1/2
: x->無限大 --------------------------
: x^1/2+100
let x^1/2 = y
l i m {x+[x+(x^1/2)]^1/2}^1/2
x->無限大 --------------------------
x^1/2+100
{y^2 + [ y^2 + y ]^1/2}^1/2
= l i m ------------------------------
y->無限大 y + 100
接下來就靠羅必達了
: l i m n
: n->無限大 --------------------------- (a>0)
: (1+a)^n
BY 羅必達
l i m n
n->無限大 --------------------------- (a>0)
(1+a)^n
1
= l i m ------------------- = 0
n->無限大 (1+a)^n * ln(1+a)
: l i m
: n->無限大 { [1/(n^2+1)^1/2]+[1/(n^2+2)^1/2]+.....
: +[1/(n^2+2n-1)^1/2]+[1/(n^2+2n)^1/2] }
2n 1
令 lim Σ --------------- = Q
n->∞ k=1 [n^2 + k]^1/2
2n 2n
lim -------------- = < Q = < lim ---------------
n->∞ [n^2 + 2n]^1/2 n->∞ [n^2 + 1]^1/2
= 2 = 2
所以 Q = 2
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