微积分求极限的题lim(x趋向于π)1+cosx/(x-π)^2

Question

微积分求极限的题
lim(x 趋向于 π) 1+cosx/(x-π)^2

Answer 1

原式=lim(x->π)[(1-cos(x-π))/(x-π)²] (应用诱导公式)
=lim(x->π)[2sin²((x-π)/2)/(x-π)²] (应用倍角公式)
=lim(x->π)[(1/2)sin²((x-π)/2)/((x-π)²/4)]
=(1/2){lim(x->π)[sin((x-π)/2)/((x-π)/2)]}²
=(1/2)*1² (令z=x-π,应用重要极限lim(z->0)(sinz/z)=1)
=1/2