حل تشریحی سوال شماره 8 حسابان 2

(0/5) ${\lim }_{x\to 1^{+}}\frac{\left[2x\right]-1}{x-1}=\frac{1}{0^{+}}=+\infty$ (الف

(0/5) ${\lim }_{x\to +\infty }\frac{2x^2-3x}{1-x^2}={\lim }_{x\to +\infty }\frac{2x^2}{-x^2}=-2$ (ب

(0/5)