Tính các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{9x + 1}}{{3x - 4}}\);

b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{7x - 11}}{{2x + 3}}\);

c) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 1} }}{x}\);

d) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 1} }}{x}\);

e) \(\mathop {\lim }\limits_{x \to 6} \frac{1}{{x - 6}}\);

f) \(\mathop {\lim }\limits_{x \to {7^ + }} \frac{1}{{x - 7}}\).

Lời giải

a) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{9x + 1}}{{3x - 4}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {9 + \frac{1}{x}} \right)}}{{x\left( {3 - \frac{4}{x}} \right)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{9 + \frac{1}{x}}}{{3 - \frac{4}{x}}} = 3\).

b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{7x - 11}}{{2x + 3}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( {7 - \frac{{11}}{x}} \right)}}{{x\left( {2 + \frac{3}{x}} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{7 - \frac{{11}}{x}}}{{2 + \frac{3}{x}}} = \frac{7}{2}\).

c) \(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 1} }}{x} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {1 + \frac{1}{{{x^2}}}} }}{x} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {1 + \frac{1}{{{x^2}}}} = 1\).

d) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 1} }}{x} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|\sqrt {1 + \frac{1}{{{x^2}}}} }}{x} = - 1\).

e) \(\mathop {\lim }\limits_{x \to 6} \frac{1}{{x - 6}} = - \infty \).

f) \(\mathop {\lim }\limits_{x \to {7^ + }} \frac{1}{{x - 7}} = + \infty \).