Câu 5: trang 142 sgk toán Đại số và giải tích 11
Tính các giới hạn sau
a. \(\mathop {\lim }\limits_{x \to 2} {{x + 3} \over {{x^2} + x + 4}}\)
b. \(\mathop {\lim }\limits_{x \to - 3} {{{x^2} + 5x + 6} \over {{x^2} + 3x}}\)
c. \(\mathop {\lim }\limits_{x \to {4^ - }} {{2x - 5} \over {x - 4}}\)
d. \(\mathop {\lim }\limits_{x \to + \infty } ( - {x^3} + {x^2} - 2x + 1)\)
e. \(\mathop {\lim }\limits_{x \to - \infty } {{x + 3} \over {3x - 1}}\)
f. \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^2} - 2x + 4} - x} \over {3x - 1}}\)
Bài Làm:
a. \(\mathop {\lim }\limits_{x \to 2} {{x + 3} \over {{x^2} + x + 4}} = {{2 + 3} \over {{2^2} + 2 + 4}} =\frac{5}{10}= {1 \over 2}\)
b. \(\mathop {\lim }\limits_{x \to - 3} {{{x^2} + 5x + 6} \over {{x^2} + 3x}} = \mathop {\lim }\limits_{x \to - 3} {{(x + 2)(x + 3)} \over {x(x + 3)}} = \mathop {\lim }\limits_{x \to - 3} {{x + 2} \over x}= {{ - 3 + 2} \over { - 3}} = {-1 \over -3} ={1 \over 3}\)
c. \(\mathop {\lim }\limits_{x \to {4^ - }} {{2x - 5} \over {x - 4}}\)
Ta có:
\(\mathop {\lim }\limits_{x \to {4^ - }} (2x - 5) = 3 > 0\)
\(\mathop {\lim }\limits_{x \to 4^-} (x - 4) = 0\)
\(\Rightarrow \mathop {\lim }\limits_{x \to {4^ - }} {{2x - 5} \over {x - 4}} = - \infty \)
d. \(\mathop {\lim }\limits_{x \to + \infty } ( - {x^3} + {x^2} - 2x + 1) = \mathop {\lim }\limits_{x \to + \infty } {x^3}( - 1 + {1 \over x} - {2 \over {{x^2}}} + {1 \over {{x^3}}}) = - \infty \)
Vì \(\mathop {\lim }\limits_{x \to + \infty } ( - 1 + {1 \over x} - {2 \over {{x^2}}} + {1 \over {{x^3}}})=-1\)
\(\mathop {\lim }\limits_{x \to + \infty } {x^3}=+\infty \)
e. \(\mathop {\lim }\limits_{x \to - \infty } {{x + 3} \over {3x - 1}} = \mathop {\lim }\limits_{x \to - \infty } {{x(1 + {3 \over x})} \over {x(3 - {1 \over x})}} = \mathop {\lim }\limits_{x \to - \infty } {{1 + {3 \over x}} \over {3 - {1 \over x}}} = {1 \over 3} \)
f. \(\mathop {\lim }\limits_{x \to - \infty } {{\sqrt {{x^2} - 2x + 4} - x} \over {3x - 1}} \)
\(= \mathop {\lim }\limits_{x \to - \infty } {{|x|\sqrt {1 - {2 \over x} + {4 \over {{x^2}}}} - x} \over {3x - 1}} \)
\(=\mathop {\lim }\limits_{x \to - \infty } {{ - x\sqrt {1 - {2 \over x} + {4 \over {{x^2}}}} - x} \over {x(3 - {1 \over x})}} \)
\(=\underset{x\rightarrow -\infty }{lim }\frac{x\left ( -\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}-1 \right )}{x(3 - {1 \over x})} \)
\(= \mathop {\lim }\limits_{x \to - \infty } {{ - \sqrt {1 - {2 \over x} + {4 \over {{x^2}}}} - 1} \over {3 - {1 \over x}}} = {{ - 2} \over 3} \).
