Nội dung bài giảng
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to 0} {{\sqrt {{x^2} + 1} - 1} \over {4 - \sqrt {{x^2} + 16} }}\) ;
b) \(\mathop {\lim }\limits_{x \to 1} {{x - \sqrt x } \over {\sqrt x - 1}}\) ;
c) \(\mathop {\lim }\limits_{x \to + \infty } {{2{x^4} + 5x - 1} \over {1 - {x^2} + {x^4}}}\) ;
d) \(\mathop {\lim }\limits_{x \to - \infty } {{x + \sqrt {4{x^2} - x + 1} } \over {1 - 2x}}\) ;
e) \(\mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {{x^2} + 1} - x} \right)\) ;
f) \(\mathop {\lim }\limits_{x \to {2^ + }} \left( {{1 \over {{x^2} - 4}} - {1 \over {x - 2}}} \right)\)
Giải:
a) 4 ; b) 1 ; c) 2; d) \({1 \over 2}\) ;
e)
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {{x^2} + 1} - x} \right) \cr
& = \mathop {\lim }\limits_{x \to + \infty } {{x\left( {{x^2} + 1 - {x^2}} \right)} \over {\sqrt {{x^2} + 1} + x}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } {x \over {x\sqrt {1 + {1 \over {{x^2}}}} + x}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } {1 \over {\sqrt {1 + {1 \over {{x^2}}}} + 1}} = {1 \over 2} \cr} \)
f)
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} \left( {{1 \over {{x^2} - 4}} - {1 \over {x - 2}}} \right) \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} {{1 - \left( {x + 2} \right)} \over {{x^2} - 4}} \cr
& = \mathop {\lim }\limits_{x \to {2^ + }} {{ - x - 1} \over {{x^2} - 4}} = - \infty \cr} \)