lim n趋向于无穷(1+e^n+派^n)^1/n,已经知道是用夹逼准则,请问怎么用
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lim n趋向于无穷(1+e^n+派^n)^1/n,已经知道是用夹逼准则,请问怎么用
![lim n趋向于无穷(1+e^n+派^n)^1/n,已经知道是用夹逼准则,请问怎么用](/uploads/image/z/18495333-45-3.jpg?t=lim+n%E8%B6%8B%E5%90%91%E4%BA%8E%E6%97%A0%E7%A9%B7%281%2Be%5En%2B%E6%B4%BE%5En%29%5E1%2Fn%2C%E5%B7%B2%E7%BB%8F%E7%9F%A5%E9%81%93%E6%98%AF%E7%94%A8%E5%A4%B9%E9%80%BC%E5%87%86%E5%88%99%2C%E8%AF%B7%E9%97%AE%E6%80%8E%E4%B9%88%E7%94%A8)
这个极限可以直接求,先进行如下变换:
(1+e^n+pi^n)^(1/n) = e^(ln((1+e^n+pi^n)^(1/n)))
= e^((1/n)*ln((1+e^n+pi^n))
然后,求下面极限:
lim_(n->+infty) (1/n)*ln((1+e^n+pi^n)
= lim_(n->+infty) (e^n+pi^n*ln(pi))/(1+e^n+pi^n) (L'Hostipal法则)
= ln(pi) (分式上下同除pi^n)
由于f(x)=e^x在实数上连续,我们得到:
lim_(n->+infty) e^((1/n)*ln((1+e^n+pi^n))
= e^(lim_(n->+infty) (1/n)*ln((1+e^n+pi^n))
= e^(ln(pi))
= pi
注:+infty表示正无穷,pi表示圆周率
(1+e^n+pi^n)^(1/n) = e^(ln((1+e^n+pi^n)^(1/n)))
= e^((1/n)*ln((1+e^n+pi^n))
然后,求下面极限:
lim_(n->+infty) (1/n)*ln((1+e^n+pi^n)
= lim_(n->+infty) (e^n+pi^n*ln(pi))/(1+e^n+pi^n) (L'Hostipal法则)
= ln(pi) (分式上下同除pi^n)
由于f(x)=e^x在实数上连续,我们得到:
lim_(n->+infty) e^((1/n)*ln((1+e^n+pi^n))
= e^(lim_(n->+infty) (1/n)*ln((1+e^n+pi^n))
= e^(ln(pi))
= pi
注:+infty表示正无穷,pi表示圆周率
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