用洛必达法则求lim x→0 tanx-x /(x²sinx)的极限lim x→0 secx/x²-1/xsinx (无穷小代换)=lim x→0 1/x²(secx-1) (洛必达法则)=lim x→0 secxtanx/2x=lim x→0 cosx/2=1/2lim x→0 secx/x²-1/xsinx (无穷小代
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![用洛必达法则求lim x→0 tanx-x /(x²sinx)的极限lim x→0 secx/x²-1/xsinx (无穷小代换)=lim x→0 1/x²(secx-1) (洛必达法则)=lim x→0 secxtanx/2x=lim x→0 cosx/2=1/2lim x→0 secx/x²-1/xsinx (无穷小代](/uploads/image/z/615045-21-5.jpg?t=%E7%94%A8%E6%B4%9B%E5%BF%85%E8%BE%BE%E6%B3%95%E5%88%99%E6%B1%82lim+x%E2%86%920+tanx-x+%2F%28x%26sup2%3Bsinx%29%E7%9A%84%E6%9E%81%E9%99%90lim+x%E2%86%920+secx%2Fx%26sup2%3B-1%2Fxsinx+%28%E6%97%A0%E7%A9%B7%E5%B0%8F%E4%BB%A3%E6%8D%A2%EF%BC%89%3Dlim+x%E2%86%920+1%2Fx%26sup2%3B%28secx-1%EF%BC%89+%EF%BC%88%E6%B4%9B%E5%BF%85%E8%BE%BE%E6%B3%95%E5%88%99%EF%BC%89%3Dlim+x%E2%86%920+secxtanx%2F2x%3Dlim+x%E2%86%920+cosx%2F2%3D1%2F2lim+x%E2%86%920+secx%2Fx%26sup2%3B-1%2Fxsinx+%28%E6%97%A0%E7%A9%B7%E5%B0%8F%E4%BB%A3)
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