Giải bài này giúp mik với mik cảm ơn

`a)`

`lim [n-1]/[2n+1]`

`= lim [n(1-1/n)]/[n(2+1/n)]`

`= lim [1-1/n]/[2+1/n]`

`= [1-0]/[2+0]`

`= 1/2`

`b)`

`lim (sqrt[n^2+2n+3]-n)`

`= lim [n^2+2n+3-n^2]/[sqrt[n^2+2n+3]+n]`

`= lim [2n+3]/[sqrt[n^2+2n+3]+n]`

`= lim [n(2+3/n)]/[n(sqrt[1+2/n+3/[n^2]]+1)]`

`= lim [2+3/n]/[sqrt[1+2/n+3/[n^2]]+1]`

`= [2+0]/[sqrt[1+0+0]+1]`

`= 1`

`c)`

`lim_[x->2] [x-2]/[x-sqrt[x+2]]`

`= lim_[x->2] [(x-2)(x+sqrt[x+2])]/[x^2-x-2]`

`= lim_[x->2] [(x-2)(x+sqrt[x+2])]/[(x+1)(x-2)]`

`= lim_[x->2] [x+sqrt[x+2]]/[x+1]`

`= [2+sqrt[2+2]]/[2+1]`

`= 4/3`

`d)`

`lim_[x->2] [sqrt[x^2+5]-3]/[2-x]`

`= lim_[x->2] [(sqrt[x^2+5]-3)(sqrt[x^2+5]+3)]/[(2-x)(sqrt[x^2+5]+3)]`

`= lim_[x->2] [x^2-4]/[(2-x)(sqrt[x^2+5]+3)]`

`= lim_[x->2] [(x-2)(x+2)]/[-(x-2)(sqrt[x^2+5]+3)]`

`= lim_[x->2] [x+2]/[-(sqrt[x^2+5]+3)]`

`= [2+2]/[-(sqrt[2^2+5]+3)]`

`= -2/3`

`e)`

`lim [3n^2+1]/[n^2-2]`

`= lim [n^2(3+1/[n^2])]/[n^2(1-2/[n^2])]`

`= lim [3+1/[n^2]]/[1-2/[n^2]]`

`= [3+0]/[1-0]`

`= 3`