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,
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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)
![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,](/uploads/image/z/2060775-63-5.jpg?t=%E5%87%A0%E9%81%93%E9%AB%98%E6%95%B0%E6%B1%82%E6%9E%81%E9%99%90%E7%9A%84%E9%A2%98%21%E6%88%91%E4%BB%AC%E5%8F%AA%E5%AD%A6%E4%BA%86%E5%A4%B9%E6%8C%A4%E5%92%8C%E5%8D%95%E8%B0%83%E6%9C%89%E7%95%8C%2C%E4%B8%A4%E4%B8%AA%E9%87%8D%E8%A6%81%E6%9E%81%E9%99%90%2C)
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