Đề bài
Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - 4}}{{x + 2}}\);
b) \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{1 - x}}\);
c) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 4x + 3}}{{x - 3}}\);
d) \(\mathop {\lim }\limits_{x \to - 2} \frac{{2 - \sqrt {x + 6} }}{{x + 2}}\);
e) \(\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {x + 1} - 1}}\);
g) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4x + 4}}{{{x^2} - 4}}\).
Phương pháp giải - Xem chi tiết
+ Sử dụng kiến thức về các phép toán về giới hạn hữu hạn của hàm số để tính: Cho \(\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L,\mathop {\lim }\limits_{x \to {x_0}} g\left( x \right) = M\): \(\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) \pm g\left( x \right)} \right] = L \pm M\), \(\mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{L}{M}\) (với \(M \ne 0\))
+ Sử dụng kiến thức về giới hạn hữu hạn cơ bản để tính: \(\mathop {\lim }\limits_{x \to {x_0}} c = c\) (với c là hằng số)
Lời giải chi tiết
a) \(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - 4}}{{x + 2}} \) \( = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{x + 2}} \) \( = \mathop {\lim }\limits_{x \to - 2} \left( {x - 2} \right)\)\( = - 2 - 2 \) \( = - 4\).
b) \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{1 - x}} \) \( = \mathop {\lim }\limits_{x \to 1} \frac{{ - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{x - 1}} \) \( = - \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + x + 1} \right)\)\( = - \left( {{1^2} + 1 + 1} \right) \) \( = - 3\);
c) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 4x + 3}}{{x - 3}} \) \( = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{x - 3}} \) \( = \mathop {\lim }\limits_{x \to 3} \left( {x - 1} \right) \) \( = 3 - 1 \) \( = 2\);
d) \(\mathop {\lim }\limits_{x \to - 2} \frac{{2 - \sqrt {x + 6} }}{{x + 2}} \) \( = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {2 - \sqrt {x + 6} } \right)\left( {2 + \sqrt {x + 6} } \right)}}{{\left( {x + 2} \right)\left( {2 + \sqrt {x + 6} } \right)}} \) \( = \mathop {\lim }\limits_{x \to - 2} \frac{{4 - x - 6}}{{\left( {x + 2} \right)\left( {2 + \sqrt {x + 6} } \right)}}\)
\( = \mathop {\lim }\limits_{x \to - 2} \frac{{ - \left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {2 + \sqrt {x + 6} } \right)}} \) \( = \mathop {\lim }\limits_{x \to - 2} \frac{{ - 1}}{{2 + \sqrt {x + 6} }} \) \( = \frac{{ - 1}}{{2 + \sqrt { - 2 + 6} }} \) \( = \frac{{ - 1}}{4}\)
e) \(\mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {x + 1} - 1}} \) \( = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt {x + 1} + 1} \right)}}{{\left( {\sqrt {x + 1} - 1} \right)\left( {\sqrt {x + 1} + 1} \right)}} \) \( = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( {\sqrt {x + 1} + 1} \right)}}{x}\)
\( = \mathop {\lim }\limits_{x \to 0} \left( {\sqrt {x + 1} + 1} \right) \) \( = \sqrt {0 + 1} + 1 \) \( = 2\);
g) \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 4x + 4}}{{{x^2} - 4}} \) \( = \mathop {\lim }\limits_{x \to 2} \frac{{{{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} \) \( = \mathop {\lim }\limits_{x \to 2} \frac{{x - 2}}{{x + 2}} \) \( = \frac{{2 - 2}}{{2 + 2}} \) \( = 0\).
