用泰勒公式计算极限,要过程

2025-04-07 10:36:08
推荐回答(2个)
回答1:

  • 泰勒公式

在数学中,泰勒公式是一个用函数在某点的信息描述其附近取值的公式。如果函数足够平滑的话,在已知函数在某一点的各阶导数值的情况之下,泰勒公式可以用这些导数值做系数构建一个多项式来近似函数在这一点的邻域中的值。泰勒公式还给出了这个多项式和实际的函数值之间的偏差。

  • 展开式

  • 极限

是微积分中的基础概念,它指的是变量在一定的变化过程中,从总的来说逐渐稳定的这样一种变化趋势以及所趋向的值(极限值)。

回答2:

(2) y → 0时, √(1+y) = 1+y/2-y^2/8+o(y^2),
因此x → 0时√(1+x^2) = 1+x^2/2-x^4/8+o(x^4),
即分子√(1+x^2)-1-x^2/2 = -x^4/8+o(x^4).
y → 0时, e^y = 1+y+o(y^2),
因此x → 0时e^(x^2) = 1+x^2+o(x^2).
又cos(x) = 1-x^2/2+o(x^2),
故e^(x^2)-cos(x) = 3x^2/2+o(x^2),
可得x^2·(e^(x^2)-cos(x)) = 3x^4/2+o(x^4).
因此x → 0时(√(1+x^2)-1-x^2/2)/(x^2·(e^(x^2)-cos(x)))
= (-x^4/8+o(x^4))/(3x^4/2+o(x^4))
= (-1/8+o(1))/(3/2+o(1))
→ -1/12.
又x → 0时, sin(x^2)/x^2 → 1,
相除即得所求极限为-1/12.

(3) 由正切差角公式, tan(arctan(x+1)-arctan(x)) = 1/(1+x+x^2),
可得arctan(x+1)-arctan(x) = arctan(1/(1+x+x^2)).
由y → 0时, arctan(y) = y+o(y).
因此x → +∞时, arctan(1/(1+x+x^2)) = 1/(1+x+x^2)+o(1/(1+x+x^2)),
x^2·arctan(1/(1+x+x^2)) = x^2/(1+x+x^2)+o(x^2/(1+x+x^2)) → 1.

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