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Tìm các giới hạn sau :
a. \(\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 1} – x} \right)\)
b. \(\mathop {\lim }\limits_{x \to 1} {{\sqrt {2x – {x^2}} – 1} \over {{x^2} – x}}\)
a. Dạng ∞ – ∞
\(\eqalign{
& \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 1} – x} \right) = \mathop {\lim }\limits_{x \to + \infty } {{{x^2} + 1 – {x^2}} \over {\sqrt {{x^2} + 1} + x}} \cr
& = \mathop {\lim }\limits_{x \to + \infty } {1 \over {\sqrt {{x^2} + 1} + x}} = 0 \cr} \)
b. Dạng \({0 \over 0}\)
\(\eqalign{
& \mathop {\lim }\limits_{x \to 1} {{\sqrt {2x – {x^2}} – 1} \over {{x^2} – x}} = \mathop {\lim }\limits_{x \to 1} {{2x – {x^2} – 1} \over {x\left( {x – 1} \right)\left( {\sqrt {2x – {x^2}} + 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 1} {{ – {{\left( {x – 1} \right)}^2}} \over {x\left( {x – 1} \right)\left( {\sqrt {2x – {x^2}} + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} {{1 – x} \over {x\left( {\sqrt {2x – {x^2}} + 1} \right)}} = 0 \cr} \)