Question

Prove that
lim┬(x□(→┬ )1^- )⁡〖f(x)≔〗 lim┬(x□(→┬ )1^- )⁡〖 (x+2)/(2x^2-3x+1)=-∞.〗

Answer 1

$$\begin{array}{l} \lim_{x \to 1 - 0} \frac{x + 2}{2x^2 - 3x + 1} = \lim_{x \to 1 - 0} \frac{x + 2}{(2x - 1)(x - 1)} = \left(\lim_{x \to 1 - 0} \frac{x + 2}{2x - 1}\right) \left(\lim_{x \to 1 - 0} \frac{1}{x - 1}\right) = \\ = \left(\frac{1 + 2}{2 - 1}\right) \left(\lim_{x \to 1 - 0} \frac{1}{x - 1}\right) = 3 \left(\lim_{x \to 1 - 0} \frac{1}{x - 1}\right) = 3(-\infty) = -\infty \end{array}$$