Question

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{4x + 3}}{{2x}}\);      

b) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{2}{{3x + 1}}\);

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

Answer 1

a: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{4+\dfrac{3}{x}}{2}=\dfrac{4}{2}=2\)

b: \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}}{3+\dfrac{1}{x}}=0\)

c: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}=1\)