Bài 4.23 trang 165 SBT đại số và giải tích 11

Đề bài

Tính các giới hiới hạn sau :

a) \(\mathop {\lim }\limits_{x \to  - 3} {{x + 3} \over {{x^2} + 2x - 3}}\) ;    b) \(\mathop {\lim }\limits_{x \to  + \infty } {{x - 1} \over {{x^2} - 1}}\);

c) \(\mathop {\lim }\limits_{x \to 5} {{x - 5} \over {\sqrt x  - \sqrt 5 }}\) ;     d) \(\mathop {\lim }\limits_{x \to  + \infty }  {{x - 5} \over {\sqrt x  + \sqrt 5 }}\) ;

e) \(\mathop {\lim }\limits_{x \to 1} {{\sqrt x  - 1} \over {\sqrt {x + 3}  - 2}}\) ;   f) \(\mathop {\lim }\limits_{x \to  + \infty } {{1 - 2x + 3{x^3}} \over {{x^3} - 9}}\) ;

g) \(\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left( {{1 \over {{x^2} + 1}} - 1} \right)\) ;

h) \(\mathop {\lim }\limits_{x \to  - \infty } {{\left( {{x^2} - 1} \right){{\left( {1 - 2x} \right)}^5}} \over {{x^7} + x + 3}}\).

a) \(\mathop {\lim }\limits_{x \to  - 3} {{x + 3} \over {{x^2} + 2x - 3}} = \mathop {\lim }\limits_{x \to  - 3} {{x + 3} \over {\left( {x - 1} \right)\left( {x + 3} \right)}} = \mathop {\lim }\limits_{x \to  - 3} {1 \over {x - 1}} = {{ - 1} \over 4}\)

b) \(\mathop {\lim }\limits_{x \to  + \infty } {{x - 1} \over {{x^2} - 1}} = \mathop {\lim }\limits_{x \to  + \infty } {{{1 \over x} - {1 \over {{x^2}}}} \over {1 - {1 \over {{x^2}}}}} = 0\)

c) \(\mathop {\lim }\limits_{x \to 5} {{x - 5} \over {\sqrt x  - \sqrt 5 }}\)

\(= \mathop {\lim }\limits_{x \to 5} {{\left( {\sqrt x  - \sqrt 5 } \right)\left( {\sqrt x  + \sqrt 5 } \right)} \over {\sqrt x  - \sqrt 5 }}\)

\(= \mathop {\lim }\limits_{x \to 5} \left( {\sqrt x  + \sqrt 5 } \right) = 2\sqrt 5 \)

d)

\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } {{x - 5} \over {\sqrt x + \sqrt 5 }} \cr 
& = \mathop {\lim }\limits_{x \to + \infty } {{1 - {5 \over x}} \over {{1 \over {\sqrt x }} + {{\sqrt 5 } \over x}}} = + \infty \cr} \)  

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

e)

\(\eqalign{
& \mathop {\lim }\limits_{x \to 1} {{\sqrt x - 1} \over {\sqrt {x + 3} - 2}} \cr 
& = \mathop {\lim }\limits_{x \to 1} {{\left( {\sqrt x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)} \over {x + 3 - 4}} \cr 
& = \mathop {\lim }\limits_{x \to 1} {{\left( {\sqrt {x - 1} } \right)\left( {\sqrt {x + 3} + 2} \right)} \over {x - 1}} \cr 
& = \mathop {\lim }\limits_{x \to 1} {{\left( {\sqrt x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)} \over {\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} \cr 
& = \mathop {\lim }\limits_{x \to 1} {{\sqrt {x + 3} + 2} \over {\sqrt x + 1}} = 2 \cr} \)

f) \(\mathop {\lim }\limits_{x \to  + \infty } {{1 - 2x + 3{x^3}} \over {{x^3} - 9}} = \mathop {\lim }\limits_{x \to  + \infty } {{{1 \over {{x^3}}} - {2 \over {{x^2}}} + 3} \over {1 - {9 \over {{x^3}}}}} = 3\)

g) 

\(\eqalign{
& \mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left( {{1 \over {{x^2} + 1}} - 1} \right) \cr 
& = \mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}.\left( {{{ - {x^2}} \over {{x^2} + 1}}} \right) \cr 
& = \mathop {\lim }\limits_{x \to 0} {{ - 1} \over {{x^2} + 1}} = - 1 \cr} \)

h)

\(\eqalign{
& \mathop {\lim }\limits_{x \to - \infty } {{\left( {{x^2} - 1} \right){{\left( {1 - 2x} \right)}^5}} \over {{x^7} + x + 3}} \cr 
& = \mathop {\lim }\limits_{x \to - \infty } {{{x^2}\left( {1 - {1 \over {{x^2}}}} \right).{x^5}{{\left( {{1 \over x} - 2} \right)}^5}} \over {{x^7} + x + 3}} \cr 
& = \mathop {\lim }\limits_{x \to - \infty } {{\left( {1 - {1 \over {{x^2}}}} \right){{\left( {{1 \over x} - 2} \right)}^5}} \over {1 + {1 \over {{x^6}}} + {3 \over {{x^7}}}}} \cr 
& = {\left( { - 2} \right)^5} = - 32 \cr}\)