求数学极限
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=lim (sinx/cosx - sinx)/x³
=lim (sinx/x) * (1-cosx)/x²
=lim (sinx/x) * 2sin²(x/2)/x² 注：半角公式。2sin²(θ/2) = 1 - cosθ
=lim (sinx/x) * 2*(1/4) *sin²(x/2) /(1/4 * x²)
=lim (sinx/x) * (1/2) * sin²(x/2) /(x/2)²
=1/2 * lim(sinx/x) * lim sin²(x/2)/(x/2)²
=1/2 * 1 * lim [sin(x/2) /(x/2)]²
=1/2 * 1 * [lim sin(x/2) /(x/2)]²
=1/2 * 1 * 1²
=1/2
=lim (sinx/x) * (1-cosx)/x²
=lim (sinx/x) * 2sin²(x/2)/x² 注：半角公式。2sin²(θ/2) = 1 - cosθ
=lim (sinx/x) * 2*(1/4) *sin²(x/2) /(1/4 * x²)
=lim (sinx/x) * (1/2) * sin²(x/2) /(x/2)²
=1/2 * lim(sinx/x) * lim sin²(x/2)/(x/2)²
=1/2 * 1 * lim [sin(x/2) /(x/2)]²
=1/2 * 1 * [lim sin(x/2) /(x/2)]²
=1/2 * 1 * 1²
=1/2
追问
可以拍照手写不。难为情。我还在解读。都解读了20分中了
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