1. **Stating the problem:** We are asked to find the limits of the function $$f(x) = (-x^3 + 2x^2) e^{-x+1}$$ as $x \to +\infty$ and as $x \to -\infty$. 2. **Recall the limit properties and behavior:** - The exponential function $e^t$ grows very fast as $t \to +\infty$ and tends to zero as $t \to -\infty$. - Polynomial terms grow slower than exponentials. 3. **Analyze the limit as $x \to +\infty$: ** Rewrite the function: $$f(x) = (-x^3 + 2x^2) e^{-x+1} = (-x^3 + 2x^2) e^{1} e^{-x}$$ As $x \to +\infty$, $e^{-x} \to 0$ very fast, while $-x^3 + 2x^2$ grows large in magnitude but polynomially. Therefore, the product tends to zero: $$\lim_{x \to +\infty} f(x) = 0$$ 4. **Analyze the limit as $x \to -\infty$: ** Rewrite the exponential term: $$e^{-x+1} = e^{1 - x} = e^{1} e^{-x}$$ As $x \to -\infty$, $-x \to +\infty$, so $e^{-x+1} \to +\infty$. The polynomial part: $$-x^3 + 2x^2$$ For large negative $x$, $-x^3$ dominates. Since $x^3$ is negative for negative $x$, $-x^3$ is positive and very large. So both parts tend to $+\infty$, thus: $$\lim_{x \to -\infty} f(x) = +\infty$$ **Final answers:** $$\boxed{\lim_{x \to +\infty} f(x) = 0, \quad \lim_{x \to -\infty} f(x) = +\infty}$$