Bài 2.6 trang 163 SBT Đại số và giải tích 11: Tính các giới hạn sau :...

Tính các giới hạn sau :

a) (mathop {lim }limits_{x o  – 3} {{x + 3} over {{x^2} + 2x – 3}}) ;                            b) (mathop {lim }limits_{x o 0} {{{{left( {1 + x} ight)}^3} – 1} over x}) ;

c) (mathop {lim }limits_{x o  + infty } {{x – 1} over {{x^2} – 1}}) ;                                d) (mathop {lim }limits_{x o 5} {{x – 5} over {sqrt x  – sqrt 5 }}) ;

e) (mathop {lim }limits_{x o  + infty }  = {{x – 5} over {sqrt x  + sqrt 5 }}) ;                      f) (mathop {lim }limits_{x o  – 2} {{sqrt {{x^2} + 5}  – 3} over {x + 2}}) ;

g) (mathop {lim }limits_{x o 1} {{sqrt x  – 1} over {sqrt {x + 3}  – 2}}) ;                               h) (mathop {lim }limits_{x o  + infty } {{1 – 2x + 3{x^3}} over {{x^3} – 9}}) ;

i) (mathop {lim }limits_{x o 0} {1 over {{x^2}}}left( {{1 over {{x^2} + 1}} – 1} ight)) ;                   j) (mathop {lim }limits_{x o  – infty } {{left( {{x^2} – 1} ight){{left( {1 – 2x} ight)}^5}} over {{x^7} + x + 3}}) ;

Giải:

a) (mathop {lim }limits_{x o  – 3} {{x + 3} over {{x^2} + 2x – 3}} = mathop {lim }limits_{x o  – 3} {{x + 3} over {left( {x – 1} ight)left( {x + 3} ight)}} = mathop {lim }limits_{x o  – 3} {1 over {x – 1}} = {{ – 1} over 4})

b) 

(eqalign{
& mathop {lim }limits_{x o 0} {{{{left( {1 + x} ight)}^3} – 1} over x} cr
& = mathop {lim }limits_{x o 0} {{left( {1 + x – 1} ight)left[ {{{left( {1 + x} ight)}^2} + left( {1 + x} ight) + 1} ight]} over x} cr
& = mathop {lim }limits_{x o 0} {{xleft[ {{{left( {1 + x} ight)}^2} + left( {1 + x} ight) + 1} ight]} over x} cr
& = mathop {lim }limits_{x o 0} left[ {{{left( {1 + x} ight)}^2} + left( {1 + x} ight) + 1} ight] = 3 cr} )

c) (mathop {lim }limits_{x o  + infty } {{x – 1} over {{x^2} – 1}} = mathop {lim }limits_{x o  + infty } {{{1 over x} – {1 over {{x^2}}}} over {1 – {1 over {{x^2}}}}} = 0)

d) (mathop {lim }limits_{x o 5} {{x – 5} over {sqrt x  – sqrt 5 }})

(= mathop {lim }limits_{x o 5} {{left( {sqrt x  – sqrt 5 } ight)left( {sqrt x  + sqrt 5 } ight)} over {sqrt x  – sqrt 5 }})

(= mathop {lim }limits_{x o 5} left( {sqrt x  + sqrt 5 } ight) = 2sqrt 5 )

e)

(eqalign{
& mathop {lim }limits_{x o + infty } {{x – 5} over {sqrt x + sqrt 5 }} cr
& = mathop {lim }limits_{x o + infty } {{1 – {5 over x}} over {{1 over {sqrt x }} + {{sqrt 5 } over x}}} = + infty cr} )  

(Vì ({1 over {sqrt x }} + {{sqrt 5 } over x} > 0) với mọi (x > 0) ).

f) 

(eqalign{
& mathop {lim }limits_{x o – 2} {{sqrt {{x^2} + 5} – 3} over {x + 2}} cr
& = mathop {lim }limits_{x o – 2} {{{x^2} + 5 – 9} over {left( {x + 2} ight)left( {sqrt {{x^2} + 5} + 3} ight)}} cr
& = mathop {lim }limits_{x o – 2} {{left( {x – 2} ight)left( {x + 2} ight)} over {left( {x + 2} ight)left( {sqrt {{x^2} + 5} + 3} ight)}} cr
& = mathop {lim }limits_{x o – 2} {{x – 2} over {sqrt {{x^2} + 5} + 3}} = {{ – 2} over 3} cr} )

g) 

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

h) (mathop {lim }limits_{x o  + infty } {{1 – 2x + 3{x^3}} over {{x^3} – 9}} = mathop {lim }limits_{x o  + infty } {{{1 over {{x^3}}} – {2 over {{x^2}}} + 3} over {1 – {9 over {{x^3}}}}} = 3)

i) 

(eqalign{
& mathop {lim }limits_{x o 0} {1 over {{x^2}}}left( {{1 over {{x^2} + 1}} – 1} ight) cr
& = mathop {lim }limits_{x o 0} {1 over {{x^2}}}.left( {{{ – {x^2}} over {{x^2} + 1}}} ight) cr
& = mathop {lim }limits_{x o 0} {{ – 1} over {{x^2} + 1}} = – 1 cr} )

j)

(eqalign{
& mathop {lim }limits_{x o – infty } {{left( {{x^2} – 1} ight){{left( {1 – 2x} ight)}^5}} over {{x^7} + x + 3}} cr
& = mathop {lim }limits_{x o – infty } {{{x^2}left( {1 – {1 over {{x^2}}}} ight).{x^5}{{left( {{1 over x} – 2} ight)}^5}} over {{x^7} + x + 3}} cr
& = mathop {lim }limits_{x o – infty } {{left( {1 – {1 over {{x^2}}}} ight){{left( {{1 over x} – 2} ight)}^5}} over {1 + {1 over {{x^6}}} + {3 over {{x^7}}}}} cr
& = {left( { – 2} ight)^5} = – 32 cr})