((1+x²)∧1/3 -1)/cosx-1 x趋向0时,求极限
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![((1+x²)∧1/3 -1)/cosx-1 x趋向0时,求极限](/uploads/image/z/5247077-5-7.jpg?t=%EF%BC%88%EF%BC%881%2Bx%26%23178%3B%EF%BC%89%E2%88%A71%2F3+-1%EF%BC%89%2Fcosx-1+x%E8%B6%8B%E5%90%910%E6%97%B6%2C%E6%B1%82%E6%9E%81%E9%99%90)
((1+x²)∧1/3 -1)/cosx-1 x趋向0时,求极限
((1+x²)∧1/3 -1)/cosx-1 x趋向0时,求极限
((1+x²)∧1/3 -1)/cosx-1 x趋向0时,求极限
x趋向0时 1-cosx等价于1/2x^2
所以x趋向0时 cosx-1等价于-1/2x^2
又x趋于0式,(1+m)^1/n-1等价于m/n
所以当x趋于时,(1+x²)∧1/3 -1等价于x^2/3
所以((1+x²)∧1/3 -1)/cosx-1 x趋向0时,极限是-2/3
解:
原式=lim(x趋向0)(1/3)(1+x²)2x/-sinx=-1/3lim(x趋向0)(6x²+2)/cosx=-1/3x2=-2/3