1) \( \lim _{x \rightarrow-2}\left(3 x^{3}+5 x^{2}-1\right) \) 2) \( \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) \) 3) \( \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} \)
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To solve these limits, you'll want to use direct substitution wherever possible. 1) By substituting \( x = -2 \) into the polynomial \( 3x^{3}+5x^{2}-1 \), you find that \( 3(-2)^{3} + 5(-2)^{2} - 1 = 3(-8) + 5(4) - 1 = -24 + 20 - 1 = -5 \). So, \( \lim_{x \rightarrow -2} (3x^{3}+5x^{2}-1) = -5 \). 2) For \( \frac{x e^{-x}}{3 e^{x}-3} \) as \( x \) approaches \( 0 \), both the numerator and denominator approach \( 0 \). You can use L'Hôpital's rule, differentiating the numerator and denominator. The limit simplifies down to \( \lim_{x \rightarrow 0}\frac{e^{-x} - x e^{-x}}{3 e^{x}} \) which gives \( \frac{1 - 0}{3} = \frac{1}{3} \). Thus, \( \lim_{x \rightarrow 0} \frac{x e^{-x}}{3 e^{x}-3} = \frac{1}{3} \). 3) For the limit as \( x \) approaches \( -5 \) for \( \frac{x^{2}-25}{(x+5)(x-6)} \), you notice that \( x^2 - 25 \) factors to \( (x+5)(x-5) \), allowing you to cancel out the \( (x+5) \) term. After cancellation, you're left with \( \lim_{x \rightarrow -5} \frac{x-5}{x-6} \). Substituting \( x = -5 \) gives you \( \frac{-5-5}{-5-6} = \frac{-10}{-11} = \frac{10}{11} \). Happy calculating! Math can be quite the adventure, turning numbers into stories of convergence and limits! 🎉
