Toán `lim_{x->oo} ( ((2x+2)(2x+1))/((x+1)(x+1)) )` …………….. 13/07/2021 By Kinsley `lim_{x->oo} ( ((2x+2)(2x+1))/((x+1)(x+1)) )` ……………..
Đáp án:`lim_{x->oo}(((2x+2)(2x+1))/((x+1)(x+1)))=4` Giải thích các bước giải: `lim_{x->oo}(((2x+2)(2x+1))/((x+1)(x+1)))` `=lim_{x->oo}((2(x+1)(2x+1))/((x+1)(x+1)))` `=lim_{x->oo}((2(2x+1))/(x+1))` `=lim_{x->oo}((4x+2)/(x+1))` `=lim_{x->oo}((4+2/x)/(1+1/x))` `=4/1=4` Trả lời
`lim_{x->oo} ( ((2x+2)(2x+1))/((x+1)(x+1)) )` `= lim_{x->oo} ((2x(1+2/(2x))(2x+1))/((x+1)(x+1)))` `= 2.lim_{x->oo} ( (x(1+2/(2x))(2x+1))/((x+1)(x+1)) )` `= 2 . lim_{x->oo} ( (2x+1)/(x+1) )` `= 2 . lim_{x->oo} ( (2+1/x)/(1+1/x) )` `= 2 . (lim_{x->oo}(2+1/x))/(lim_{x->oo}( 1 + 1/x))` `= 2 . 2/1` `= 4` Trả lời
Đáp án:`lim_{x->oo}(((2x+2)(2x+1))/((x+1)(x+1)))=4`
Giải thích các bước giải:
`lim_{x->oo}(((2x+2)(2x+1))/((x+1)(x+1)))`
`=lim_{x->oo}((2(x+1)(2x+1))/((x+1)(x+1)))`
`=lim_{x->oo}((2(2x+1))/(x+1))`
`=lim_{x->oo}((4x+2)/(x+1))`
`=lim_{x->oo}((4+2/x)/(1+1/x))`
`=4/1=4`
`lim_{x->oo} ( ((2x+2)(2x+1))/((x+1)(x+1)) )`
`= lim_{x->oo} ((2x(1+2/(2x))(2x+1))/((x+1)(x+1)))`
`= 2.lim_{x->oo} ( (x(1+2/(2x))(2x+1))/((x+1)(x+1)) )`
`= 2 . lim_{x->oo} ( (2x+1)/(x+1) )`
`= 2 . lim_{x->oo} ( (2+1/x)/(1+1/x) )`
`= 2 . (lim_{x->oo}(2+1/x))/(lim_{x->oo}( 1 + 1/x))`
`= 2 . 2/1`
`= 4`