lim(x→-∞){xarctan[e^(-x)]}/√(x²-xsinx1)

lim(x→-∞）e^xsinxlim(x→-∞）e^xsinxlim(x→-∞）e^xsinxe^xsin(x),当x趋于负无穷时,e^x趋于e^(负无穷）=1/e^(负无穷）,因为分母趋于无穷大,

可以一下以下三道求极限的题吗?1.lim(x->无穷大）[x^3ln(x+1/x-1)-2x^2];答案：2/32.lim(x->无穷大）[xarctan(1/x)]^(x^2)答案：e^83.lim

lim(x→+∞)(x+e^x)^(1/x)lim(x→+∞)(x+e^x)^(1/x)lim(x→+∞)(x+e^x)^(1/x)解答如下：在matlab软件中求得的解如下：f=(x+exp(x))

lim(x-1)[e^(1/x)-1],x→+∞lim(x-1)[e^(1/x)-1],x→+∞lim(x-1)[e^(1/x)-1],x→+∞lim(x-1)[e^(1/x)-1]=limx[e^(

lim（e＾x）cosx=x→+∞lim（e＾x）cosx=x→+∞lim（e＾x）cosx=x→+∞x→+∞则e^x→+∞而cosx在[-1,1]震荡,即有界所以e^x*cosx→∞极限不存在无穷大

Lim(x/e)^((x-e)^-1),x→eLim(x/e)^((x-e)^-1),x→eLim(x/e)^((x-e)^-1),x→eLim(x/e)^((x-e)^-1)=lim(1+(x-e)

Lim(1+x)e^x/e^x-1x→-∞Lim(1+x)e^x/e^x-1x→-∞Lim(1+x)e^x/e^x-1x→-∞Lim(1+x)e^x/e^x-1x→-∞=Lim(1+x)e^x/e^x

lim(x→0)[e^x-e^(2x)]/xlim(x→0)[e^x-e^(2x)]/xlim(x→0)[e^x-e^(2x)]/xlim(x→0)[e^x-e^(2x)]/x洛必达法则求导得,原式=

lim(1-1/3x)^xx→∞求极限lim(1+1/x)^x=elim(1-1/3x)^xx→∞求极限lim(1+1/x)^x=elim(1-1/3x)^xx→∞求极限lim(1+1/x)^x=el

求极限.😂谢谢了!当x→+∞时,lime∧x=?lime∧-x=?lime∧1=?当x→-∞时,lime∧x=?lime∧-x=?lime∧1=?求极限.😂谢谢了!当x

lim[cosx/(e^x+e^-x]x→+∞,求极限,要过程哦.lim[cosx/(e^x+e^-x]x→+∞,求极限,要过程哦.lim[cosx/(e^x+e^-x]x→+∞,求极限,要过程哦.l

极限运算lim(x→∞)cosx/(e^1/x+e^-1/x)极限运算lim(x→∞)cosx/(e^1/x+e^-1/x)极限运算lim(x→∞)cosx/(e^1/x+e^-1/x)点击放大,再点

lim(x→0)[e^x-e^(-x)]/sinx=?lim(x→0)[e^x-e^(-x)]/sinx=?lim(x→0)[e^x-e^(-x)]/sinx=?用罗比达法则,即分子分母同时求导!（0

lim(x→∞)e^x/[(1+1/x)^x^2]求极限,lim(x→∞)e^x/[(1+1/x)^（x^2）]lim(x→∞)e^x/[(1+1/x)^x^2]求极限,lim(x→∞)e^x/[(1

解极限题目limx→∞(e^x+x)^9/x用5/x解极限题目limx→∞(e^x+x)^9/x用5/x解极限题目limx→∞(e^x+x)^9/x用5/x

f(x)=(2+e^x)/1+e^2x)+|x|sin1/x求1)lim[x→+∞]f(x);2)lim[x→-∞]f(x);3)lim[x→∞]f(x)f(x)=(2+e^x)/1+e^2x)+|x

lim(x→0)e^x-x-1/x^2lim(x→0)e^x-x-1/x^2lim(x→0)e^x-x-1/x^2lim (e^x-1)/x = 1x→0这个是由重要极限

求极限lim(tanx)^tan2x,x→∏/4真是让人着急,数学式子不好输入.原题是用洛必达解得,是1^∞型.lim(tanx)^tan2x=e^limtan2xlntanx=e^limlntanx

求极限lim1-2x/e^x^2,x→+∞＝0原题求极限x→+∞,limf（x）＝lim1-2x/e^x^2＝lim－2／2e^x^2＝0说明：e^x^2表示e的x次方的2次方.请问：lim1-2x/

请教用洛必达法则求极限lim┬(x→+∞)x^2*e^(1/x^2请教用洛必达法则求极限lim┬(x→+∞)x^2*e^(1/x^2请教用洛必达法则求极限lim┬(x→+∞)x^2*e^(1/x^2l