1.limx-》0(x*【1/x】)的极限2.limx-》{x^2/(x^2-1)}^x的极限3.limn-》无穷{(1+x)(1+x^2).(1+x^2n)}(x绝对值小于1)的极限4已知lim-》1f(x)/(x-1)^2=-1,证明在x=1点的某去心领域内有f(x)《05.n趋于无穷是,2^n为无
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![1.limx-》0(x*【1/x】)的极限2.limx-》{x^2/(x^2-1)}^x的极限3.limn-》无穷{(1+x)(1+x^2).(1+x^2n)}(x绝对值小于1)的极限4已知lim-》1f(x)/(x-1)^2=-1,证明在x=1点的某去心领域内有f(x)《05.n趋于无穷是,2^n为无](/uploads/image/z/2682894-30-4.jpg?t=1.limx-%E3%80%8B0%EF%BC%88x%2A%E3%80%901%2Fx%E3%80%91%EF%BC%89%E7%9A%84%E6%9E%81%E9%99%902.limx-%E3%80%8B%7Bx%5E2%2F%28x%5E2-1%29%7D%5Ex%E7%9A%84%E6%9E%81%E9%99%903.limn-%E3%80%8B%E6%97%A0%E7%A9%B7%7B%281%2Bx%29%281%2Bx%5E2%29.%281%2Bx%5E2n%29%7D%28x%E7%BB%9D%E5%AF%B9%E5%80%BC%E5%B0%8F%E4%BA%8E1%29%E7%9A%84%E6%9E%81%E9%99%904%E5%B7%B2%E7%9F%A5lim-%E3%80%8B1f%28x%29%2F%28x-1%29%5E2%3D-1%2C%E8%AF%81%E6%98%8E%E5%9C%A8x%3D1%E7%82%B9%E7%9A%84%E6%9F%90%E5%8E%BB%E5%BF%83%E9%A2%86%E5%9F%9F%E5%86%85%E6%9C%89f%EF%BC%88x%EF%BC%89%E3%80%8A05.n%E8%B6%8B%E4%BA%8E%E6%97%A0%E7%A9%B7%E6%98%AF%2C2%5En%E4%B8%BA%E6%97%A0)
1.limx-》0(x*【1/x】)的极限2.limx-》{x^2/(x^2-1)}^x的极限3.limn-》无穷{(1+x)(1+x^2).(1+x^2n)}(x绝对值小于1)的极限4已知lim-》1f(x)/(x-1)^2=-1,证明在x=1点的某去心领域内有f(x)《05.n趋于无穷是,2^n为无
1.limx-》0(x*【1/x】)的极限
2.limx-》{x^2/(x^2-1)}^x的极限
3.limn-》无穷{(1+x)(1+x^2).(1+x^2n)}(x绝对值小于1)的极限
4已知lim-》1f(x)/(x-1)^2=-1,证明在x=1点的某去心领域内有f(x)《0
5.n趋于无穷是,2^n为无穷大
.刚刚学..完全不懂呀.,
1.limx-》0(x*【1/x】)的极限2.limx-》{x^2/(x^2-1)}^x的极限3.limn-》无穷{(1+x)(1+x^2).(1+x^2n)}(x绝对值小于1)的极限4已知lim-》1f(x)/(x-1)^2=-1,证明在x=1点的某去心领域内有f(x)《05.n趋于无穷是,2^n为无
第一题估计【1/x】是取整,要不太简单了.
用夹逼x(1+1/x)
(1)我看错了,第一题是取整:
1/x-1<[1/x]<1/x,所以(1-x)
所以原极限=1
(2)lim(x->∞)[x^2/(x^2-1)]^x=lim(x->∞)[1+1/(x^2-1)]^[(x^2-1)/x](因为(x^2-1)/x=[x-(...
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(1)我看错了,第一题是取整:
1/x-1<[1/x]<1/x,所以(1-x)
所以原极限=1
(2)lim(x->∞)[x^2/(x^2-1)]^x=lim(x->∞)[1+1/(x^2-1)]^[(x^2-1)/x](因为(x^2-1)/x=[x-(1/x)]->x)
=lim(x->∞){[1+1/(x^2-1)]^(x^2-1)}^(1/x)=lim(x->∞)e^1/x=e^0=1
(3)(1+x)(1+x^2)....(1+x^2n)左乘一个(1-x),运用平方差公式,等于(1-x^4n)
原极限=lim(n->∞)[(1-x^4n)/(1-x)]=1/(1-x)
(4)lim(x->1)[f(x)/(x-1)^2]=-1,可得在x=1点的某去心领域内,|f(x)/(x-1)^2-(-1)|<ε
即:|f(x)+(x-1)^2|<ε*(x-1)^2,因为在x=1点的某去心领域,所以|x-1|<√δ(δ为某一正常数)
-δ*ε
(5)当n->+∞,时,对任意正实数N>0,存在正实数M>log2(N+ε),都有n>M,
所以|2^n-N|>|N+ε-N|=ε,所以n->∞时,2^n->∞。
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