c) \( \lim _{x \rightarrow -\infty} \frac{(x+2)^{4}-(x-1)^{4}}{(2 x+3)^{3}}, \mathrm{R}: \frac{3}{2} \)

To solve the limit \( \lim _{x \rightarrow -\infty} \frac{(x+2)^{4}-(x-1)^{4}}{(2 x+3)^{3}} \), we can start by simplifying the expression in the numerator. 1. **Numerator Simplification**: We have \( (x + 2)^4 - (x - 1)^4 \). Expanding both terms gives us: \[ (x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16 \] \[ (x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 \] Subtracting these we get: \[ (x + 2)^4 - (x - 1)^4 = (x^4 + 8x^3 + 24x^2 + 32x + 16) - (x^4 - 4x^3 + 6x^2 - 4x + 1) \] \[ = 12x^3 + 18x^2 + 36x + 15 \] 2. **Denominator Simplification**: The denominator is: \[ (2x + 3)^3 \] Expanding this, we have: \[ (2x + 3)^3 = 8x^3 + 36x^2 + 54x + 27 \] Now we can rewrite the limit: \[ \lim_{x \to -\infty} \frac{12x^3 + 18x^2 + 36x + 15}{8x^3 + 36x^2 + 54x + 27} \] 3. **Divide by Highest Power**: We divide both the numerator and denominator by \( x^3 \): \[ = \lim_{x \to -\infty} \frac{12 + \frac{18}{x} + \frac{36}{x^2} + \frac{15}{x^3}}{8 + \frac{36}{x} + \frac{54}{x^2} + \frac{27}{x^3}} \] 4. **Evaluate the Limit**: As \( x \to -\infty \), the terms \( \frac{18}{x}, \frac{36}{x^2}, \frac{15}{x^3}, \frac{36}{x}, \frac{54}{x^2}, \) and \( \frac{27}{x^3} \) approach 0. Thus, we are left with: \[ \frac{12}{8} = \frac{3}{2} \] Therefore, the final result is: \[ \lim _{x \rightarrow -\infty} \frac{(x+2)^{4}-(x-1)^{4}}{(2 x+3)^{3}} = \frac{3}{2} \]