(x+sin∧2x+tanx)/(sinx+x∧2)在x→0时的极限
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![(x+sin∧2x+tanx)/(sinx+x∧2)在x→0时的极限](/uploads/image/z/2982954-66-4.jpg?t=%28x%2Bsin%E2%88%A72x%2Btanx%29%2F%28sinx%2Bx%E2%88%A72%29%E5%9C%A8x%E2%86%920%E6%97%B6%E7%9A%84%E6%9E%81%E9%99%90)
(x+sin∧2x+tanx)/(sinx+x∧2)在x→0时的极限
(x+sin∧2x+tanx)/(sinx+x∧2)在x→0时的极限
(x+sin∧2x+tanx)/(sinx+x∧2)在x→0时的极限
是0:0型,用洛必达法则求导得:
(1+2sinxcosx+1+sec^2x)/(cosx+2x)
=(1+1+1)/1
=3
∫sin^2x(1+tanx)dx
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已知tanx=2,则sin^2x+1=?
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化简(sin^2x+2sinxcosx)/1+tanx