Câu 44 trang 167 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 } xsqrt {{{2{x^3} + x} over {{x^5} – {x^2} + 3}}} )

b.  (mathop {lim }limits_{x o – infty } {{left| x ight| + sqrt {{x^2} + x} } over {x + 10}})

c.  (mathop {lim }limits_{x o + infty } {{sqrt {2{x^4} + {x^2} – 1} } over {1 – 2x}})

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

Giải:

a. Với (x < 0), ta có :

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

Do đó :  (mathop {lim }limits_{x o – infty } xsqrt {{{2{x^3} + x} over {{x^5} – {x^2} + 3}}} = – sqrt 2 )

b.

(eqalign{
& mathop {lim }limits_{x o – infty } {{left| x ight|+sqrt {{x^2} + x} } over {x + 10}} = mathop {lim }limits_{x o – infty } {{left| x ight| + left| x ight|sqrt {1 + {1 over x}} } over {x + 10}} cr
& = mathop {lim }limits_{x o – infty } {{ – x – xsqrt {1 + {1 over x}} } over {x + 10}} = mathop {lim }limits_{x o – infty } {{ – 1 – sqrt {1 + {1 over x}} } over {1 + {{10} over x}}} cr &= – 2 cr} )

c.

(eqalign{
& mathop {lim }limits_{x o + infty } {{sqrt {2{x^4} + {x^2} – 1} } over {1 – 2x}} = mathop {lim }limits_{x o + infty } {{{x^2}sqrt {2 + {1 over {{x^2}}} – {1 over {{x^4}}}} } over {xleft( {{1 over x} – 2} ight)}} cr
& = mathop {lim }limits_{x o + infty } x{{sqrt {2 + {1 over {{x^2}}} – {1 over {{x^4}}}} } over {{1 over x} – 2}} = – infty cr
& ext{vì},mathop {lim }limits_{x o + infty } x = + infty , ext{và},mathop {lim }limits_{x o + infty } {{sqrt {2 + {1 over {{x^2}}} – {1 over {{x^4}}}} } over {{1 over x} – 2}} = – {{sqrt 2 } over 2} < 0 cr} )

d.

(eqalign{
& mathop {lim }limits_{x o – infty } left( {sqrt {2{x^2} + 1} + x} ight) = mathop {lim }limits_{x o – infty } {{2{x^2} + x – {x^2}} over {sqrt {2{x^2} + x} – x}} cr
& = mathop {lim }limits_{x o – infty } {{xleft( {x + 1} ight)} over { – xleft( {sqrt {2 + {1 over x}} + 1} ight)}} cr &= mathop {lim }limits_{x o – infty } – {{x + 1} over {sqrt {2 + {1 over x} + 1} }} = + infty cr
& ext{vì },mathop {lim }limits_{x o – infty } left( { – x – 1} ight) = + infty cr} )