求lim[x趨于無(wú)窮]{(x^3)*ln[(x+1)/(x-1)]-2x^2},答案為2/3,

數(shù)學(xué)人氣：480 ℃時(shí)間：2020-02-05 21:15:28

優(yōu)質(zhì)解答

lim[x趨于無(wú)窮]{(x^3)*ln[(x+1)/(x-1)]-2x^2}
=lim[x趨于無(wú)窮]{(x^2){xln[(x+1)/(x-1)]-2}
=lim[x趨于無(wú)窮]{xln[1+2/(x-1)]-2}/[x^(-2)]
=lim[x趨于無(wú)窮]{ln[1+2/(x-1)]+x[1+2/(x-1)]^(-1)[-2/(x-1)^2]}/[-2x^(-3)]
=lim[x趨于無(wú)窮]{-2x/(x-1)^2[-2x^(-3)]
=2/3極限性質(zhì)，復(fù)合函數(shù)求極限，極限符號(hào)可以通過(guò)表達(dá)式。你一點(diǎn)都不笨，只說(shuō)明你很認(rèn)真。堅(jiān)持！