lim(4/(x^2-4)+1/(x-2)= (n→2)
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/08 23:12:16
lim(4/(x^2-4)+1/(x-2)= (n→2)
算出来lim=-1/(x+2)答案是-1/4,我觉得是0
是lim(4/(x^2-4)-1/(x-2)= (n→2)
算出来lim=-1/(x+2)答案是-1/4,我觉得是0
是lim(4/(x^2-4)-1/(x-2)= (n→2)
![lim(4/(x^2-4)+1/(x-2)= (n→2)](/uploads/image/z/16431705-9-5.jpg?t=lim%284%2F%28x%5E2-4%29%2B1%2F%28x-2%29%3D+%EF%BC%88n%E2%86%922%EF%BC%89)
如果题没错的话应该是无穷大吧,按答案来看题应该是lim(4/(x^2-4)-1/(x-2))= (x→2)这样吧
既然改了题,那就对了.4/(x^2-4)可以写为[1/(x-2)]-[1/(x+2)],接下来抵消一项,直接带入x=2就ok了
既然改了题,那就对了.4/(x^2-4)可以写为[1/(x-2)]-[1/(x+2)],接下来抵消一项,直接带入x=2就ok了
lim(x→∞)1+2+3+…+n/(n+2)(n+4)=?
求极限,lim(x->0) (1-2sinx)^(3/x)lim(n->+∞) (n!-4^n) / (6+ln(n)+
求极限lim (n→+∞)ln(1+x^2)/ln(1+x^4)
lim(x→∞)(n/2^n)=lim(x→∞)(1/(ln2*2^n)) 这个是咋么算出来的,
求极限lim(x→∞)(1/n+2/n+3/n..+n/n)
微积分lim(根号n^4+n+1)(3n+4)= (n->∞)lim{[(5x^2+1/3x-1)*sinx^-1]+(
极限题lim(1+1/2+1/4+.+1/2^n) n→无穷lim((x^2 -1)/(x^2 +3))^(x^2 +1
求极限:lim(x→无穷)(2^n-7^n)/(2^n+7^n-1)=?
求f(x)=lim(n→∞)[x^(n+2)-x^n]/[x^n+x^(-n-1)]的间断点集齐类型
请老师回答问题,lim(n趋于无穷大)(1^n+2^n+3^n)1/n次方=?lim(x趋于无穷大)sin2x/x=?
求极限:lim(n->∞)(2x^2-3x-4)/(1+x^4)^1/2
求两道极限计算题:1)lim(x->1):(x^n-1)/(x-1) (n属于自然数集,x^n表示x的n次方)2)lim