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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) \( \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} \)

Ask by Huang Henry. in the United Kingdom
Jan 23,2025

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Answer

The limits are: 1. \( \lim _{x \rightarrow -2}\left(3 x^{3}+5 x^{2}-1\right) = -5 \) 2. \( \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) = \frac{1}{3} \) 3. \( \lim _{x \rightarrow -5} \frac{x^{2}-25}{(x+5)(x-6)} = \frac{10}{11} \)

Solution

Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow -2}\left(3x^{3}+5x^{2}-1\right)\) - step1: Rewrite the expression: \(\lim _{x\rightarrow -2}\left(3x^{3}\right)+\lim _{x\rightarrow -2}\left(5x^{2}\right)-\lim _{x\rightarrow -2}\left(1\right)\) - step2: Calculate: \(-24+20-1\) - step3: Calculate: \(-4-1\) - step4: Calculate: \(-5\) Calculate the limit \( \lim_{x\rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow -5}\left(\frac{x^{2}-25}{\left(x+5\right)\left(x-6\right)}\right)\) - step1: Reduce the fraction: \(\lim _{x\rightarrow -5}\left(\frac{x-5}{x-6}\right)\) - step2: Rewrite the expression: \(\frac{\lim _{x\rightarrow -5}\left(x-5\right)}{\lim _{x\rightarrow -5}\left(x-6\right)}\) - step3: Evaluate: \(\frac{-10}{\lim _{x\rightarrow -5}\left(x-6\right)}\) - step4: Evaluate: \(\frac{-10}{-11}\) - step5: Reduce the fraction: \(\frac{10}{11}\) Calculate the limit \( \lim_{x\rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) \). Evaluate the limit by following steps: - step0: Evaluate using L'Hopital's rule: \(\lim _{x\rightarrow 0}\left(\frac{xe^{-x}}{3e^{x}-3}\right)\) - step1: Rewrite the expression: \(\lim _{x\rightarrow 0}\left(\frac{1}{3e^{x}-3}\times xe^{-x}\right)\) - step2: Calculate: \(\lim _{x\rightarrow 0}\left(\frac{1}{3}e^{-x}\right)\) - step3: Calculate: \(\lim _{x\rightarrow 0}\left(\frac{1}{3}\times 1\right)\) - step4: Multiply the terms: \(\lim _{x\rightarrow 0}\left(\frac{1}{3}\right)\) - step5: Calculate: \(\frac{1}{3}\) Here are the results for the limits you requested: 1) For the limit \( \lim _{x \rightarrow -2}\left(3 x^{3}+5 x^{2}-1\right) \): \[ \lim _{x \rightarrow -2}\left(3 x^{3}+5 x^{2}-1\right) = -5 \] 2) For the limit \( \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) \): \[ \lim _{x \rightarrow 0}\left(\frac{x e^{-x}}{3 e^{x}-3}\right) = \frac{1}{3} \] 3) For the limit \( \lim _{x \rightarrow -5} \frac{x^{2}-25}{(x+5)(x-6)} \): \[ \lim _{x \rightarrow -5} \frac{x^{2}-25}{(x+5)(x-6)} = \frac{10}{11} \] If you have any further questions or need additional calculations, feel free to ask!

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Bonus Knowledge

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 \(\frac{10}{11}\).

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