1. **State the problem:** Calculate the limit $$\lim_{x\to +\infty} \frac{1 - 6x^3}{2x^3 - x^2 + 5x - 3} + \lim_{x \to -\infty} \frac{1}{|x| + 1}$$ 2. **Recall the rules:** - For limits involving rational functions as $x \to \pm \infty$, divide numerator and denominator by the highest power of $x$ in the denominator. - For absolute value, $|x| = -x$ when $x \to -\infty$. 3. **Calculate the first limit:** $$\lim_{x\to +\infty} \frac{1 - 6x^3}{2x^3 - x^2 + 5x - 3}$$ Divide numerator and denominator by $x^3$: $$\lim_{x\to +\infty} \frac{\frac{1}{x^3} - 6}{2 - \frac{1}{x} + \frac{5}{x^2} - \frac{3}{x^3}}$$ As $x \to +\infty$, terms with $\frac{1}{x^n} \to 0$, so: $$\frac{0 - 6}{2 - 0 + 0 - 0} = \frac{-6}{2} = -3$$ 4. **Calculate the second limit:** $$\lim_{x \to -\infty} \frac{1}{|x| + 1}$$ Since $x \to -\infty$, $|x| = -x$, so: $$\lim_{x \to -\infty} \frac{1}{-x + 1}$$ Rewrite as: $$\lim_{x \to -\infty} \frac{1}{-x + 1} = \lim_{x \to -\infty} \frac{1}{-x(1 - \frac{1}{x})}$$ As $x \to -\infty$, $-x \to +\infty$ and $1 - \frac{1}{x} \to 1$, so denominator $\to +\infty$. Therefore, the limit is: $$0$$ 5. **Add the two limits:** $$-3 + 0 = -3$$ **Final answer:** $$\boxed{-3}$$