Câu 23 trang 152 SGK Đại số và Giải tích 11 Nâng cao

Đề bài

Bài 23. Tìm các giới hạn sau :

a.  \(\mathop {\lim }\limits_{x \to 2} \left( {3{x^2} + 7x + 11} \right)\)

b.  \(\mathop {\lim }\limits_{x \to 1} {{x - {x^3}} \over {\left( {2x - 1} \right)\left( {{x^4} - 3} \right)}}\)

c.  \(\mathop {\lim }\limits_{x \to 0} x\left( {1 - {1 \over x}} \right)\)

d.  \(\mathop {\lim }\limits_{x \to 9} {{\sqrt x - 3} \over {9x - {x^2}}}\)

e.  \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} - 4} \right|\)

f.  \(\mathop {\lim }\limits_{x \to 2} \sqrt {{{{x^4} + 3x - 1} \over {2{x^2} - 1}}} \)

Hướng dẫn giải

a.  \(\eqalign{& \mathop {\lim }\limits_{x \to 2} \left( {3{x^2} + 7x + 11} \right) = \mathop {\lim }\limits_{x \to 2} 3{x^2} + \mathop {\lim }\limits_{x \to 2} 7x + \mathop {\lim }\limits_{x \to 2} 11 \cr & = {3.2^2} + 7.2 + 11 = 37 \cr} \)

b.  \(\mathop {\lim }\limits_{x \to 1} {{x - {x^3}} \over {\left( {2x - 1} \right)\left( {{x^4} - 3} \right)}} = {0 \over { - 2}} = 0\)

c.  \(\mathop {\lim }\limits_{x \to 0} x\left( {1 - {1 \over x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {x - 1} \right) = - 1\)

d.  \(\mathop {\lim }\limits_{x \to 9} {{\sqrt x - 3} \over {9x - {x^2}}} = \mathop {\lim }\limits_{x \to 9} {{\sqrt x - 3} \over { - x\left( {x - 9} \right)}} = - \mathop {\lim }\limits_{x \to 9} {1 \over {x\left( {\sqrt x + 3} \right)}} = - {1 \over {54}}\)

e.  \(\mathop {\lim }\limits_{x \to \sqrt 3 } \left| {{x^2} - 4} \right| = 1\)

f.  \(\mathop {\lim }\limits_{x \to 2} \sqrt {{{{x^4} + 3x - 1} \over {2{x^2} - 1}}} = \sqrt {{{{2^4} + 3.2 - 1} \over {{{22}^2} - 1}}} = \sqrt 3 \)