lim arctan e^x
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5.lim(e^x^2)-1/arctanx-05.lim(e^x^2)-1/arctanx-05.lim(e^x^2)-1/arctanx-0我猜了一下是不是下面这个极限:exp(x^2)-1lim
lim(x→0)arctan(sinx/x)lim(x→0)arctan(sinx/x)lim(x→0)arctan(sinx/x)原式=arctan[lim(x→0)(sinx/x)]=arctan
lim(x→+∞)(2/πarctanX)^Xlim(x→+∞)(2/πarctanX)^Xlim(x→+∞)(2/πarctanX)^X假设a=(2/πarctanX)^Xlna=x*ln(2/πa
limx(arctanx-π/2)x–>∞limx(arctanx-π/2)x–>∞limx(arctanx-π/2)x–>∞lim(x–>+∞)(arctanx-π/2)/(1/x)=lim(x–>
Lim(x→a)arctan(x-a)/(x-a)Lim(x→a)arctan(x-a)/(x-a)Lim(x→a)arctan(x-a)/(x-a)令t=x-a原式=lim(t->0)arctant
lim(arctanX)/x,x趋近于0,怎么解?lim(arctanX)/x,x趋近于0,怎么解?lim(arctanX)/x,x趋近于0,怎么解?洛必达法则,lim=(上下求导)(1/(1+x^2
lim(arctan1/x+arccot1/x)(xlim(arctan1/x+arccot1/x)(xlim(arctan1/x+arccot1/x)(x极限不存在.1.当x趋向于0+的时候,1/x
急求!lim(x-arctan(cosx))/(x+arctan(cosx)),x趋向于无穷大急求!lim(x-arctan(cosx))/(x+arctan(cosx)),x趋向于无穷大急求!lim
lim(x→0)arctan1/x等于多少lim(x→0)arctan1/x等于多少lim(x→0)arctan1/x等于多少这个极限不存在,因为它的左右极限存在但不相等:lim(x->0-)arct
一高数题lim(x趋向于无穷大)arctan(x/3)是多少一高数题lim(x趋向于无穷大)arctan(x/3)是多少一高数题lim(x趋向于无穷大)arctan(x/3)是多少lim(x趋向于无穷
求lime^(1/x^2)*arctan((x^2+2-1)/(x+1)*(x-2))当x趋近于0,-1,2的值求lime^(1/x^2)*arctan((x^2+2-1)/(x+1)*(x-2))当
limarctan[1/(x^2-1)]求左右极限limarctan[1/(x^2-1)]求左右极限limarctan[1/(x^2-1)]求左右极限左边->1,负无穷,右边->1正无穷,所以分别为-
cot(arctan(e^x))等于什么?cot(arctan(e^x))等于什么?cot(arctan(e^x))等于什么?arctanx+arctan(1/x)=π/2因此cot(arctan(e
关于洛必达法则的高数题(求极限):@@arctanx1/x^2-1/3@lim(---------------)只知道答案是e,thx~@x->0x-sinx@@arctanx1/x^2-1/3@li
x→0limarctan(sinx/x)可以写成arctanlim(sinx/x)吗?那limln(sinx/x)可以写成这种形式吗?为什么?x→0limarctan(sinx/x)可以写成arcta
arctan(e^x)+arctan(e^-x)=x的计算过程arctan(e^x)+arctan(e^-x)=x的计算过程arctan(e^x)+arctan(e^-x)=x的计算过程设f(x)=a
arctan(e^x)+arctan(e^-x)为什么等于二分之π?arctan(e^x)+arctan(e^-x)为什么等于二分之π?arctan(e^x)+arctan(e^-x)为什么等于二分之
求下列不定积分∫(arctane^x)/(e^x)dx求下列不定积分∫(arctane^x)/(e^x)dx求下列不定积分∫(arctane^x)/(e^x)dx∫arctan(e^x)/(e^x)d
求∫arctan(e^x)/(e^x)dx?求∫arctan(e^x)/(e^x)dx?求∫arctan(e^x)/(e^x)dx?a=e^xx=lnadx=da/a所以原式=∫arctana*da/
∫(0→1)arctan(e^x)/e^xdx∫(0→1)arctan(e^x)/e^xdx∫(0→1)arctan(e^x)/e^xdx给你思路吧--具体自己算好么化成arctan(e^x)/e^2