Câu 32 trang 159 SGK Đại số và Giải tích 11 Nâng cao, Tìm các giới hạn sau :...

Tìm các giới hạn sau :

a.  (mathop {lim }limits_{x o + infty } oot 3 of {{{2{x^5} + {x^3} – 1} over {left( {2{x^2} – 1} ight)left( {{x^3} + x} ight)}}} )

b.  (mathop {lim }limits_{x o – infty } {{2left| x ight| + 3} over {sqrt {{x^2} + x + 5} }})

c.  (mathop {lim }limits_{x o – infty } {{sqrt {{x^2} + x} + 2x} over {2x + 3}})

d.  (mathop {lim }limits_{x o + infty } left( {x + 1} ight)sqrt {{x over {2{x^4} + {x^2} + 1}}} )

Giải

a.  (mathop {lim }limits_{x o + infty } oot 3 of {{{2{x^5} + {x^3} – 1} over {left( {2{x^2} – 1} ight)left( {{x^3} + x} ight)}}} = mathop {lim }limits_{x o + infty } oot 3 of {{{2 + {1 over {{x^2}}} – {1 over {{x^5}}}} over {left( {2 – {1 over {{x^2}}}} ight)left( {1 + {1 over {{x^2}}}} ight)}}} = 1)

b.

(eqalign{
& mathop {lim }limits_{x o – infty } {{2left| x ight| + 3} over {sqrt {{x^2} + x + 5} }} = mathop {lim }limits_{x o – infty } {{2left| x ight| + 3} over {left| x ight|sqrt {1 + {1 over x} + {5 over {{x^2}}}} }} cr
& = mathop {lim }limits_{x o – infty } {{ – 2x + 3} over { – xsqrt {1 + {1 over x} + {5 over {{x^2}}}} }} =mathop {lim }limits_{x o  – infty } {{2 – {3 over x}} over {sqrt {1 + {1 over x} + {5 over {{x^2}}}} }}= 2 cr} )

c.  ({x^2} + x ge 0 Leftrightarrow x le – 1, ext{ hoặc },x ge 0)

Với mọi (x ≤ -1),  (x e – {3 over 2})

({{sqrt {{x^2} + x} + 2x} over {2x + 3}} = {{left| x ight|sqrt {1 + {1 over x}} + 2x} over {2x + 3}} = {{ – xsqrt {1+ {1 over x}} + 2x} over {2x + 3}} = {{ – sqrt {1 + {1 over x}} + 2} over {2 + {3 over x}}})

Do đó  (mathop {lim }limits_{x o – infty } {{sqrt {{x^2} + x} + 2x} over {2x + 3}} =mathop {lim }limits_{x o – infty }{{ – sqrt {1 + {1 over x}} + 2} over {2 + {3 over x}}}= {1 over 2})

d.

(eqalign{
& mathop {lim }limits_{x o + infty } left( {x + 1} ight)sqrt {{x over {2{x^4} + {x^2} + 1}}} = mathop {lim }limits_{x o + infty } sqrt {{{x{{left( {x + 1} ight)}^2}} over {2{x^4} + {x^2} + 1}}} cr
& = mathop {lim }limits_{x o + infty } sqrt {{{{1 over x} + {2 over {{x^2}}} + {1 over {{x^3}}}} over {2 + {1 over {{x^2}}} + {1 over {{x^4}}}}}} = 0 cr} )