Question

高数求极限题2道，要求方法越多越好，避开洛比达法则，等价无穷小尽量别用，用了也无所谓

只要击中要点即可，不需要全部过程

Answer 1

点击放大、再点击再放大：

Answer 2

①
⑴等价无穷小量：
a^x -1 ~ xlna

lim(x->+∞) x^2[a^(1/x) - a^(1/(x+1))]

=lim(x->+∞) x^2 * [a^[(1/x)-1/(x+1)] - 1 ]

=lim(x->+∞) x^2* a^(1/(x+1)) *[a^(1/(x^2+x) - 1]

=lim(x->+∞) x^2 a^(1/(x+1)) [lna*1/(x^2+x)]

=lna

⑵ 重要极限(多此一举哈！）

lim(x->+∞) x^2[a^(1/x) - a^(1/(x+1))]

=lim(x->+∞) [x^2lna/(x^2+x)] *a^(1/(x+1))* [e^[lna/(x^2+x)] - 1]/[lna/(x^2+x)]
令：e^[lna/(x^2+x)] - 1 = t ；lna/(x^2+x) = ln(1+t)
=lim(x->+∞) [x^2lna/(x^2+x)] *a^(1/(x+1)) lim(t->0) t/ln(1+t)

=lna*lim(t->0) 1/ln[(1+t)^(1/t)]

= lna

②
⑴等价无穷小量：

lim(x->0) {(1+4x)^(1/3)*(1+3x)^(1/4) - 1 }/x

=lim(x->0) {[(1+4x)^(1/3)-1]*[(1+3x)^(1/4) - 1] +[(1+4x)^(1/3)-1]+[(1+3x)^(1/4)-1]}/x

=lim(x->0)[(1+4x)^(1/3)-1]*[(1+3x)^(1/4) - 1]/x + lim(x->0)[(1+4x)^(1/3)-1]/x + lim(x->0)[(1+3x)^(1/4)-1]}/x

=lim(x->0) 4x/3* 3x/4*1/x + lim(x->0) 4x/3*1/x + lim(x->0) 3x/4*1/x

= 0 + 4/3 + 3/4

= 25/12

⑵ Taylor公式：
(1+x)^a = 1 + ax + a(a-1)/2! x^2 + a(a-1)(a-2)/3! x^3 + o(x^3)

lim(x->0) {(1+4x)^(1/3)*(1+3x)^(1/4) - 1 }/x

=lim(x->0) {(1+4x/3+o(x))*(1+3x/4+o(x)) - 1 }/x

=lim(x->0) {(1+4x/3+o(x))*(1+3x/4+o(x)) - 1 }/x

=lim(x->0) {[4x/3+3x/4+o(x))}/x

= 25/12

Answer 3

①
⑴等价无穷小量：
a^x -1 ~ xlna
lim(x->+∞) x^2[a^(1/x) - a^(1/(x+1))]
=lim(x->+∞) x^2 * [a^[(1/x)-1/(x+1)] - 1 ]
=lim(x->+∞) x^2* a^(1/(x+1)) *[a^(1/(x^2+x) - 1]
=lim(x->+∞) x^2 a^(1/(x+1)) [lna*1/(x^2+x)]
=lna
⑵ 重要极限
lim(x->+∞) x^2[a^(1/x) - a^(1/(x+1))]
=lim(x->+∞) [x^2lna/(x^2+x)] *a^(1/(x+1))* [e^[lna/(x^2+x)] - 1]/[lna/(x^2+x)]
令：e^[lna/(x^2+x)] - 1 = t ；lna/(x^2+x) = ln(1+t)
=lim(x->+∞) [x^2lna/(x^2+x)] *a^(1/(x+1)) lim(t->0) t/ln(1+t)
=lna*lim(t->0) 1/ln[(1+t)^(1/t)]
= lna
②
⑴等价无穷小量：
lim(x->0) {(1+4x)^(1/3)*(1+3x)^(1/4) - 1 }/x
=lim(x->0) {[(1+4x)^(1/3)-1]*[(1+3x)^(1/4) - 1] +[(1+4x)^(1/3)-1]+[(1+3x)^(1/4)-1]}/x
=lim(x->0)[(1+4x)^(1/3)-1]*[(1+3x)^(1/4) - 1]/x + lim(x->0)[(1+4x)^(1/3)-1]/x + lim(x->0)[(1+3x)^(1/4)-1]}/x
=lim(x->0) 4x/3* 3x/4*1/x + lim(x->0) 4x/3*1/x + lim(x->0) 3x/4*1/x
= 0 + 4/3 + 3/4
= 25/12
⑵ Taylor公式：
(1+x)^a = 1 + ax + a(a-1)/2! x^2 + a(a-1)(a-2)/3! x^3 + o(x^3)
lim(x->0) {(1+4x)^(1/3)*(1+3x)^(1/4) - 1 }/x
=lim(x->0) {(1+4x/3+o(x))*(1+3x/4+o(x)) - 1 }/x
=lim(x->0) {(1+4x/3+o(x))*(1+3x/4+o(x)) - 1 }/x
=lim(x->0) {[4x/3+3x/4+o(x))}/x
= 25/12

Answer 4

不用洛比达法则，不用等价无穷小你做什么极限
剩下给你的工具只有拉格朗日中值定理和泰勒公式
这两个公式还非常不好用。
有时候能化成级数求解，但是这两道题中只有第二题可以用级数解。

其他有些情况能判断极限是0这样的特殊值的时候我们才不用那些技巧

Answer 5

用泰勒公式展开，这用方法最稳妥