Câu 31 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 – sqrt 2 } {{{x^3} + 2sqrt 2 } over {{x^2} – 2}})

b.  (mathop {lim }limits_{x o 3} {{{x^4} – 27x} over {2{x^2} – 3x – 9}})

c.  (mathop {lim }limits_{x o – 2} {{{x^4} – 16} over {{x^2} + 6x + 8}})

d.  (mathop {lim }limits_{x o {1^ – }} {{sqrt {1 – x} + x – 1} over {sqrt {{x^2} – {x^3}} }})

Giải:

a. Ta có:

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

b.

(eqalign{
& mathop {lim }limits_{x o 3} {{{x^4} – 27x} over {2{x^2} – 3x – 9}} = mathop {lim }limits_{x o 3} {{xleft( {x – 3} ight)left( {{x^2} + 3x + 9} ight)} over {left( {x – 3} ight)left( {2x + 3} ight)}} cr
& = mathop {lim }limits_{x o 3} {{xleft( {{x^2} + 3x + 9} ight)} over {2x + 3}} = 9 cr} )

c.

(eqalign{
& mathop {lim }limits_{x o – 2} {{{x^4} – 16} over {{x^2} + 6x + 8}} = mathop {lim }limits_{x o – 2} {{left( {{x^2} – 4} ight)left( {{x^2} + 4} ight)} over {left( {x + 2} ight)left( {x + 4} ight)}} cr
& = mathop {lim }limits_{x o – 2} {{left( {x – 2} ight)left( {{x^2} + 4} ight)} over {x + 4}} = – 16 cr} )

d.

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