几道高数求极限的题!我们只学了夹挤和单调有界,两个重要极限,
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/23 17:33:32
几道高数求极限的题!
我们只学了夹挤和单调有界,两个重要极限,
![](http://img.wesiedu.com/upload/3/ed/3edf038c1e19f8f333aee71c884ed814.jpg)
我们只学了夹挤和单调有界,两个重要极限,
![](http://img.wesiedu.com/upload/3/ed/3edf038c1e19f8f333aee71c884ed814.jpg)
1.(1)lim n=∞(n→∞);lim a^n=∞(n→∞,a>1);lim n/a^n=lim 1/(a^n*lna)=0
(2)limx->无穷{(1+1/n+1/n2)^1/(1/n+1/n2)}1+1/n=e
3.lim(x->1)(1-x)tan(πx/2)
=lim(y->0)[y*tan(π/2-πy/2)] (用y=1-x代换)
=lim(y->0)[y*ctan(πy/2)]
=lim(y->0)[y*cos(πy/2)/sin(πy/2)]
=lim(y->0){[(πy/2)/sin(πy/2)]*[(2/π)*cos(πy/2)]}
={lim(y->0)[(πy/2)/sin(πy/2)]}*{lim(y->0)[(2/π)*cos(πy/2)]}
=1*(2/π) (应用重要极限lim(x->0(sinx/x)=1)
=2/π
4.limx->0f(x)=2
limx->2f(x)=limx->2x^+1=5
先这么多吧
(2)limx->无穷{(1+1/n+1/n2)^1/(1/n+1/n2)}1+1/n=e
3.lim(x->1)(1-x)tan(πx/2)
=lim(y->0)[y*tan(π/2-πy/2)] (用y=1-x代换)
=lim(y->0)[y*ctan(πy/2)]
=lim(y->0)[y*cos(πy/2)/sin(πy/2)]
=lim(y->0){[(πy/2)/sin(πy/2)]*[(2/π)*cos(πy/2)]}
={lim(y->0)[(πy/2)/sin(πy/2)]}*{lim(y->0)[(2/π)*cos(πy/2)]}
=1*(2/π) (应用重要极限lim(x->0(sinx/x)=1)
=2/π
4.limx->0f(x)=2
limx->2f(x)=limx->2x^+1=5
先这么多吧