Bài 13 trang 180 Toán Đại số và giải tích 11: Tính các giới hạn sau...

Bài 13. Tính các giới hạn sau

a) (mathop {lim }limits_{x o  – 2} {{6 – 3x} over {sqrt {2{x^2} + 1} }})

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

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

d) (mathop {lim }limits_{x o {1^ – }} (x + {x^2} + … + {x^n} – {n over {1 – x}});n in {N^*})

e) (mathop {lim }limits_{x o  + infty } {{2x – 1} over {x – 3}})

f) (mathop {lim }limits_{x o  – infty } {{x + sqrt {4{x^2} – 1} } over {2 – 3x}})

g) (mathop {lim }limits_{x o  – infty } ( – 2{x^3} + {x^2} – 3x + 1))

Trả lời:

a) (mathop {lim }limits_{x o  – 2} {{6 – 3x} over {sqrt {2{x^2} + 1} }} = {{6 – 3( – 2)} over {sqrt {2{{( – 2)}^2} + 1} }} = {{12} over 3} = 4)

b)

(eqalign{
& mathop {lim }limits_{x o 2} {{x – sqrt {3x – 2} } over {{x^2} – 4}} cr
& = mathop {lim }limits_{x o 2} {{(x – sqrt {x – 2} )(x + sqrt {3x – 2} )} over {({x^2} – 4)(x + sqrt {3x – 2} )}} cr
& = mathop {lim }limits_{x o 2} {{{x^2} – 3x + 2} over {({x^2} – 4)(x + sqrt {3x – 2} )}} cr
& = mathop {lim }limits_{x o 2} {{(x – 2)(x – 1)} over {(x – 2)(x + 2)(x + sqrt {3x – 2)} }} cr
& = mathop {lim }limits_{x o 2} {{x – 1} over {(x + 2)(x + sqrt {3x – 2} )}} cr
& = {{2 – 1} over {(2 + 2)(2 + sqrt {3.2 – 2} )}} = {1 over {16}} cr} )

c) Ta có:

+) (mathop {lim }limits_{x o {2^ + }} ({x^2} – 3x + 1) = 4 – 6 + 1 =  – 1)

+) (left{ matrix{
x – 2 > 0 hfill cr
mathop {lim }limits_{x o {2^ + }} (x – 2) = 0 hfill cr} ight.)

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

d) Ta có:

(eqalign{
& mathop {lim }limits_{x o 1^-} (x + {x^2} + … + {x^n} – {n over {1 – x}}) = – infty cr
& left{ matrix{
1 – x > 0,forall x < 1 hfill cr
mathop {lim }limits_{x o {1^ – }} (1 – x) = 0 hfill cr} ight. cr} )

+ Suy ra: (mathop {lim }limits_{x o {1^ – }} {n over {1 – x}} =  + infty )

+ Do đó: (mathop {lim }limits_{x o {1^ – }} (x + {x^2} + … + {x^n} – {n over {1 – x}}) =  – infty )

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

f)

(eqalign{
& mathop {lim }limits_{x o – infty } {{x + sqrt {4{x^2} – 1} } over {2 – 3x}} cr
& = mathop {lim }limits_{x o – infty } {{x + |x|sqrt {4 – {1 over {{x^2}}}} } over {2 – 3x}} cr
& = mathop {lim }limits_{x o – infty } {{x – xsqrt {4 – {1 over {{x^2}}}} } over {2 – 3x}} cr
& = mathop {lim }limits_{x o – infty } {{x(1 – sqrt {4 – {1 over {{x^2}}}} )} over {x({2 over x} – 3)}} cr
& = mathop {lim }limits_{x o – infty } {{1 – sqrt {4 – {1 over {{x^2}}}} } over {{2 over x} – 3}} cr
& = {{1 – sqrt 4 } over { – 3}} = {1 over 3} cr} )

g)

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