1. **Statement of the problem:** Given the function $$f(x) = 2x + 1 - xe^{-x}$$, calculate the limits $$\lim_{x \to +\infty} f(x)$$ and $$\lim_{x \to -\infty} f(x)$$. 2. **Recall the limit properties and behavior of exponential functions:** - As $$x \to +\infty$$, $$e^{-x} \to 0$$ because the exponential decays rapidly. - As $$x \to -\infty$$, $$e^{-x} = e^{|x|} \to +\infty$$ because the exponent becomes very large positive. 3. **Calculate $$\lim_{x \to +\infty} f(x)$$:** $$f(x) = 2x + 1 - xe^{-x}$$ As $$x \to +\infty$$, $$e^{-x} \to 0$$, so $$xe^{-x} \to x \cdot 0 = 0$$. Therefore, $$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (2x + 1 - 0) = \lim_{x \to +\infty} (2x + 1) = +\infty$$. 4. **Calculate $$\lim_{x \to -\infty} f(x)$$:** Rewrite the term: $$xe^{-x} = x e^{|x|}$$ where $$|x| = -x$$ since $$x < 0$$. As $$x \to -\infty$$, $$e^{|x|} = e^{-x} \to +\infty$$ and $$x \to -\infty$$. The product $$x e^{|x|}$$ behaves like $$(-\infty) \times (+\infty)$$ which tends to $$-\infty$$ because $$x$$ is negative and $$e^{|x|}$$ grows faster than any polynomial. So, $$f(x) = 2x + 1 - x e^{-x} = 2x + 1 - x e^{|x|}$$ Since $$- x e^{|x|} \to +\infty$$ (because $$x e^{|x|} \to -\infty$$), the dominant term is $$- x e^{|x|}$$ which tends to $$+\infty$$. Therefore, $$\lim_{x \to -\infty} f(x) = +\infty$$. **Final answers:** $$\lim_{x \to +\infty} f(x) = +\infty$$ $$\lim_{x \to -\infty} f(x) = +\infty$$