Bài 2.6 trang 163 Sách bài tập (SBT) Đại số và giải tích 11
Tính các giới hạn sau : ...
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})
