lim (2n-1)/(2^n)=?n=>无穷大
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lim(2n-1)/(2^n)=?n=>无穷大lim(2n-1)/(2^n)=?n=>无穷大lim(2n-1)/(2^n)=?n=>无穷大0.当n>=10时2^(n/2)>2n-1（此式的证明可用构造
lim (2n-1)/(2^n)=?n=>无穷大
lim (2n-1)/(2^n)=?n=>无穷大
lim (2n-1)/(2^n)=?n=>无穷大
0.
当n>=10时2^(n/2)>2n-1
（此式的证明可用构造函数f（x）=2^(x/2)-2x+1,则由f(10)>0,
f`(x)=2^x*ln2/2-2>0 ,(x>10时),f(x)为增函数；所以f(x)>0,(x>10)）
则当n=>无穷大时,0
等于0
用罗必塔法则
lim(1-1/n)^(n^2)=?
lim(1-1/n^2)^n=?
lim (2n-1)/(2^n)=?n=>无穷大
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