Đề bài

Cho \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 4}}{{x - 1}} = 2\). Tính a) \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\)                                                    b) \(\mathop {\lim }\limits_{x \to 1} 3f\left( x \right)\)

Lời giải chi tiết

a) Giả sử \(\mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 4} \right) = L \ne 0\). Khi đó \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 4}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 4} \right).\mathop {\lim }\limits_{x \to 1} \frac{1}{{x - 1}}\).Ta nhận thấy \(\mathop {\lim }\limits_{x \to 1} \frac{1}{{x - 1}} =  + \infty \) hoặc \(\mathop {\lim }\limits_{x \to 1} \frac{1}{{x - 1}} =  - \infty \), nên \(\mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 4} \right)\mathop {\lim }\limits_{x \to 1} \frac{1}{{x - 1}}\) không thể bằng 2.Do vậy \(\mathop {\lim }\limits_{x \to 1} \left( {f\left( x \right) - 4} \right) = 0 \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 4\).b) Theo câu a, ta có \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = 4\), suy ra: \(\mathop {\lim }\limits_{x \to 1} 3f\left( x \right) = \mathop {\lim }\limits_{x \to 1} 3.\mathop {\lim }\limits_{x \to 1} f\left( x \right) = 3.4 = 12\).