Bài 6 trang 133 sgk đại số 11

Bài 6 trang 133 sgk đại số 11

Tính:

Bài 6. Tính:

(eqalign{
& a)mathop {lim }limits_{x o + infty } ({x^4} - {x^2} + x - 1) cr
& b)mathop {lim }limits_{x o - infty } ( - 2{x^3} + 3{x^2} - 5) cr
& c)mathop {lim }limits_{x o - infty } (sqrt {{x^2} - 2x + 5}) cr
& d)mathop {lim }limits_{x o + infty } {{sqrt {{x^2} + 1} + x} over {5 - 2x}} cr} )

Giải:

(eqalign{
& a)mathop {lim }limits_{x o + infty } ({x^4} - {x^2} + x - 1) = mathop {lim }limits_{x o + infty } {x^4}left( {1 - {1 over {{x^2}}} + {1 over {{x^3}}} - {1 over {{x^4}}}} ight) = + infty cr
& b)mathop {lim }limits_{x o - infty } ( - 2{x^3} + 3{x^2} - 5) = mathop {lim }limits_{x o - infty } {x^3}left( { - 2 + {1 over x} - {5 over {{x^2}}}} ight) = + infty cr
& c)mathop {lim }limits_{x o - infty } (sqrt {{x^2} - 2x + 5} ) = mathop {lim }limits_{x o - infty } |x|sqrt {1 - {2 over x} + {5 over {{x^2}}}} = + infty cr
& d)mathop {lim }limits_{x o + infty } {{sqrt {{x^2} + 1} + x} over {5 - 2x}} = mathop {lim }limits_{x o + infty } {{xleft( {sqrt {1 + {1 over {{x^2}}}} + 1} ight)} over {5 - 2x}} = mathop {lim }limits_{x o + infty } {{left( {sqrt {1 + {1 over {{x^2}}}} + 1} ight)} over {{5 over x} - 2}} = - 1 cr} )

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