Question

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to  - 2} \left( {{x^2} - 7x + 4} \right)\)  

b) \(\mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{{x^2} - 9}}\)                                    

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{3 - \sqrt {x + 8} }}{{x - 1}}\)

Answer 1

a: \(\lim\limits_{x\rightarrow-2}x^2-7x+4=\left(-2\right)^2-7\cdot\left(-2\right)+4=22\)

b: \(\lim\limits_{x\rightarrow3}\dfrac{x-3}{x^2-9}=\lim\limits_{x\rightarrow3}\dfrac{1}{x+3}=\dfrac{1}{3+3}=\dfrac{1}{6}\)

c: \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-x-8}{3+\sqrt{x+8}}\cdot\dfrac{1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{-1}{3+\sqrt{x+8}}\)

\(=-\dfrac{1}{6}\)