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 tackle the limit \( \lim _{x \rightarrow-2}(3 x^{3}+5 x^{2}-1) \), we can simply substitute \( x = -2 \). This gives us \( 3(-2)^3 + 5(-2)^2 - 1 = 3(-8) + 5(4) - 1 = -24 + 20 - 1 = -5 \). So, the limit is \(-5\). For \( \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) \), we have an indeterminate form \( \frac{0}{0} \). Applying L'Hôpital's rule, differentiate the numerator and the denominator. The derivative of the numerator \( x e^{-x} \) is \( e^{-x} - x e^{-x} \) and the derivative of \( 3 e^{x} - 3 \) is \( 3 e^{x} \). Evaluating the limit after applying L'Hôpital's gives us \( \lim_{x \to 0} \frac{e^{-0} - 0 \cdot e^{0}}{3 e^{0}} = \frac{1 - 0}{3} = \frac{1}{3} \). Hence, the limit is \(\frac{1}{3}\). For \( \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} \), we note that at \( x = -5 \) both the numerator and denominator approach zero, giving another indeterminate form \( \frac{0}{0} \). Factor the numerator as \( (x - 5)(x + 5) \). The expression simplifies to \( \frac{(x - 5)}{(x - 6)} \) when \( x \neq -5 \). Now, substituting \( x = -5 \) computes directly: \( \frac{-10}{-11} = \frac{10}{11} \). Thus, the limit is \(\frac{10}{11}\).
