大一数学 极限问题

\u5927\u4e00\u6570\u5b66\u6781\u9650\u9898\u2026

1\u3001
\u89e3\uff1a
\u5f53x\u2192\u221e\u65f6\uff0csin(1/x)
+
cos(3/\u221ax)\u21921;
\u6307\u65702x\u2192\u221e\uff0c\u56e0\u6b64\u8be5\u6781\u9650\u4e3a1^\u221e
\u578b\u6781\u9650\uff0c\u5e94\u6309\u7167\u8be5\u7c7b\u578b\u6781\u9650\u7684\u6c42\u89e3\u601d\u8def\uff0c\u5148\u5c06\u5e95\u6570\u4e2d\u5206\u89e3\u51fa1+0(x)\u7684\u5f62\u5f0f,\u5176\u4e2d0(x)\u662f\u4e00\u4e2a\u5173\u4e8ex\u7684\u51fd\u6570\uff0c\u6ee1\u8db3(x\u21920)
\u65f60(x)\u7684\u6781\u9650\u4e3a0\uff0c\u7136\u540e\u5c06\u6781\u9650\u5316\u4e3alim
(1+0(x))^[(1/0(x)
*f(x)]\u5f62\u5f0f\uff0c\u8fdb\u4e00\u6b65\u5316\u4e3alim
{(1+0(x))^[1/0(x)]}^f(x)\u7684\u5f62\u5f0f\uff0c\u6839\u636e\u91cd\u8981\u6781\u9650(x\u21920)lim
(1+x)^(1/x)
=e,
\u539f\u6781\u9650\u5316\u4e3a(x\u21920)lime^f(x)\u7684\u5f62\u5f0f\uff0c\u518d\u6c42\u89e3
f(x)\u7684\u6781\u9650\u5c31\u53ef\u4ee5\u89e3\u51fa\u6574\u4f53\u6781\u9650\u503c\u4e86\u3002
\u5f53x\u2192\u221e\u65f6\uff0c1/x
\u21920\uff0c\u4ee4y=1/x\uff0c\u539f\u5f0f\u5316\u4e3a\uff1a
\u539f\u5f0f=
(y\u21920)lim[
sin(y)+cos(3\u221ay)]^(2/y)
=
(y\u21920)lim[
sin(y)+cos(3\u221ay)]^(2/y)
=
(y\u21920)lim[1+(
sin(y)+cos(3\u221ay)-1)]^(2/y)
//\u4e0b\u4e00\u6b65\u5316\u4e3a(x\u21920)lim
(1+x)^(1/x)\u5f62\u5f0f
=
(y\u21920)lim[1+(
sin(y)+cos(3\u221ay)-1)]^[(1/(
sin(y)+cos(3\u221ay)-1))*(2(
sin(y)+cos(3\u221ay)-1)/y)
=
(y\u21920)lim{[1+(
sin(y)+cos(3\u221ay)-1)]^[1/(
sin(y)+cos(3\u221ay)-1)]}^[2siny/y
-2(1-cos(3\u221ay))/y]
\u5229\u7528\u534a\u89d2\u516c\u5f0f\uff1a1-cos2x
=
2sin²x,
\u6709
2(1-cos(3\u221ay))/y
=
4
sin²(3\u221ay/2)/y
=
4[sin(3/2*\u221ay)/(
3/2*\u221ay)]²
*9/4
=
9[sin(3/2*\u221ay)/(
3/2*\u221ay)]²
\u56e0\u6b64\uff0c\u539f\u5f0f=
(y\u21920)lim[
sin(y)+cos(3\u221ay)]^(2/y)
=
(y\u21920)lim[
sin(y)+cos(3\u221ay)]^(2/y)
=
(y\u21920)lim[1+(
sin(y)+cos(3\u221ay)-1)]^(2/y)
//\u4e0b\u4e00\u6b65\u5316\u4e3a(x\u21920)lim
(1+x)^(1/x)\u5f62\u5f0f
=
(y\u21920)lim[1+(
sin(y)+cos(3\u221ay)-1)]^[(1/(
sin(y)+cos(3\u221ay)-1))*(2(
sin(y)+cos(3\u221ay)-1)/y)
=
(y\u21920)lim{[1+(
sin(y)+cos(3\u221ay)-1)]^[1/(
sin(y)+cos(3\u221ay)-1)]}^{2siny/y
-9[sin(3/2*\u221ay)/(
3/2*\u221ay)]²}
=
e^(2-9)
=
e^(-7)
2\u3001
\u601d\u8def\u540c1\uff0c\u5148\u5316\u4e3a1^\u221e\u7684\u6807\u51c6\u5f62\u5f0f(x\u21920)lim
(1+x)^[(1/x)*f(x)]\u5f62\u5f0f\uff0c\u518d\u6c42\u89e3\u6307\u6570\u90e8\u5206\u7684\u6781\u9650
\u539f\u5f0f=
(x\u21920)lim[1+(a^x+b^x+c^x-3)/3]^(1/x)
=
(x\u21920)lim[1+(a^x+b^x+c^x-3)/3]^
[3/(a^x+b^x+c^x-3)
*
(a^x+b^x+c^x-3)/(3x)]
=
(x\u21920)lim{[1+(a^x+b^x+c^x-3)/3]^
[3/(a^x+b^x+c^x-3)]}^[(a^x+b^x+c^x-3)/(3x)]
=
(x\u21920)lim
e^{1/3*
[(a^x-1)/(3x)+(b^x-1)/(3x)+(c^x-1)/(3x)]}
\u7531\u4e8e
(x\u21920)lim(a^x-1)/x
=
(x\u21920)lim[e^(x*lna)-1)]/x
\u6211\u4eec\u77e5\u9053
x\u21920\u65f6\uff0ce^x
-1
\u7684\u7b49\u4ef7\u65e0\u7a77\u5c0f\u4e3ax\uff0c\u56e0\u6b64e^(x*lna)-1\u7b49\u4ef7\u65e0\u7a77\u5c0f\u4e3ax*lna
\u56e0\u6b64(x\u21920)lim(a^x-1)/x
=
(x\u21920)lim[e^(x*lna)-1)]/x
=
(x\u21920)lim[e^(x*lna)-1)]/(x*lna)
*lna
=
lna
\u539f\u5f0f
=
(x\u21920)lim
e^{1/3*
[(a^x-1)/(3x)+(b^x-1)/(3x)+(c^x-1)/(3x)]}
=
e^[1/3
*(lna+lnb+lnc)]
=
∛(abc)

\uff081\uff09
\u5f53x\u8d8b\u4e8e0\u65f6\uff0c(1+x)^a~1+ax\uff0ce^x-1~x\uff0csinx~x
\u6240\u4ee5\u539f\u5f0f=lim [(1/2)xsinx]/(x²)\uff08\u5c06\u5206\u5b50\u7684\u6839\u53f7\u770b\u505a1/2\u6b21\u65b9\uff0c\u7136\u540e\u5229\u7528\u7b2c\u4e00\u4e2a\u7b49\u4ef7\u5f0f\uff0c\u5206\u6bcd\u7528\u7b2c\u4e8c\u4e2a\uff09
=lim (x²)/(2x²)\uff08\u5229\u7528\u7b2c\u4e09\u4e2a\u7b49\u4ef7\uff09
=1/2

\uff082\uff09
x\u8d8b\u4e8e\u65e0\u7a77\u65f6\uff0c1/x\u8d8b\u4e8e0
\u6240\u4ee5e^(1/x)\u8d8b\u4e8ee^0=1
\u6240\u4ee5 lim e^(1/x)-1=1-1=0

1.1/2，上下同除以X方
2,4倍根2/9 ，上下同时有理化
3.0，无穷小×有界函数还是无穷小
4.0，罗比达法则

4个提示
1、分子分母因式分解后消去（x-1)
2、分子分母同乘以根号（x-2)+根号2
3、x→∞时，分子arctgx→正负π/2，为有界量，而分母为无穷大量，有界量除以无穷大，所以极限为0
4、分子分母同除以sinx，利用x→0时，lim x/sinx=1来解.

1、式子可化为（X-1）/（2X+1），所以答案为2

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