Question

高数极限lim[1-cosx(cos2x)^(1/2)]/x^2求解,在线等,快快

Answer 1

解：lim1-cosx(cos2x)^(1/2)/x^2
=lim1-cosx(cos2x)^(1/2)*1+cosx(cos2x)^(1/2)/{x^21+cosx(cos2x)^(1/2)}
=lim1-cosx^2cos2x/{x^21+cosx(cos2x)^(1/2)}
=lim1-cosx^2(cosx^2-sinx^2)/{x^21+cosx(cos2x)^(1/2)}
=limcosx^2+sinx^2+cosx^2sinx^2-cosx^4/{x^21+cosx(cos2x)^(1/2)}
=lim
sinx^2(1+2cosx^2)/{x^21+cosx(cos2x)^(1/2)}
=limsinx^2/x^2*lim(1+2cosx^2)/1+cosx(cos2x)^(1/2)
=1*lim(1+2cosx^2)/1+cosx(cos2x)^(1/2)
x=0代入：原式=3/2
解题要点就是分子分母同时×1+cosx(cos2x)^(1/2)分子就丛哪可以有理化，而档游1+cosx(cos2x)^(1/渗蠢码2)=0这类式子在x=0时总是确定的值，而不会趋于0，所以是可以计算的，而不用再按罗比塔法则求导