\( ( \lim _ { x \rightarrow 0 } \frac { - \ln ( 1 + 7 ( e ^ { - x } - 1 ) ) } { x } ) ^ { 6 } + ( ( \prod _ { = 1 } ^ { 3 } k ) ^ { 4 } + - 2 \operatorname { mng } e ^ { - x } ) \)

Evaluate the limit by following steps: - step0: Evaluate using L'Hopital's rule: \(\lim _{x\rightarrow 0}\left(\frac{-\ln{\left(1+7\left(e^{-x}-1\right)\right)}}{x}\right)\) - step1: Add the terms: \(\lim _{x\rightarrow 0}\left(\frac{-\ln{\left(-6+7e^{-x}\right)}}{x}\right)\) - step2: Use the L'Hopital's rule: \(\lim _{x\rightarrow 0}\left(\frac{\frac{d}{dx}\left(-\ln{\left(-6+7e^{-x}\right)}\right)}{\frac{d}{dx}\left(x\right)}\right)\) - step3: Find the derivative: \(\lim _{x\rightarrow 0}\left(\frac{-\frac{7}{6e^{x}-7}}{1}\right)\) - step4: Simplify the expression: \(\lim _{x\rightarrow 0}\left(\frac{-7}{6e^{x}-7}\right)\) - step5: Rewrite the expression: \(\frac{-7}{\lim _{x\rightarrow 0}\left(6e^{x}-7\right)}\) - step6: Calculate: \(\frac{-7}{-1}\) - step7: Divide the terms: \(7\) Calculate or simplify the expression \( (1*2*3)^{4} \). Calculate the value by following steps: - step0: Calculate: \(\left(1\times 2\times 3\right)^{4}\) - step1: Multiply the terms: \(6^{4}\) - step2: Evaluate the power: \(1296\) Calculate or simplify the expression \( 1296 - 2 * e^{-x} \). Simplify the expression by following steps: - step0: Solution: \(1296-2e^{-x}\) Calculate or simplify the expression \( (7)^{6} + (1296 - 2e^{-x}) \). Simplify the expression by following steps: - step0: Solution: \(7^{6}+\left(1296-2e^{-x}\right)\) - step1: Remove the parentheses: \(7^{6}+1296-2e^{-x}\) Let's break down the problem step by step. 1. **Limit Calculation**: We need to evaluate the limit: \[ \lim_{x \rightarrow 0} \frac{-\ln(1 + 7(e^{-x} - 1))}{x} \] As \( x \) approaches 0, \( e^{-x} \) approaches 1. Therefore, \( e^{-x} - 1 \) approaches 0. This means we can simplify the expression inside the logarithm: \[ 1 + 7(e^{-x} - 1) \rightarrow 1 + 7(0) = 1 \] Thus, the limit becomes: \[ \lim_{x \rightarrow 0} \frac{-\ln(1)}{x} = \lim_{x \rightarrow 0} \frac{0}{x} = 0 \] However, we need to evaluate it more carefully using L'Hôpital's Rule since it is in the indeterminate form \( \frac{0}{0} \). Applying L'Hôpital's Rule: \[ \text{Differentiate the numerator and denominator:} \] The derivative of the numerator \( -\ln(1 + 7(e^{-x} - 1)) \) is: \[ -\frac{7(-e^{-x})}{1 + 7(e^{-x} - 1)} = \frac{7e^{-x}}{1 + 7(e^{-x} - 1)} \] The derivative of the denominator \( x \) is \( 1 \). Thus, we have: \[ \lim_{x \rightarrow 0} \frac{7e^{-x}}{1 + 7(e^{-x} - 1)} = \frac{7 \cdot 1}{1 + 7(0)} = 7 \] 2. **Product Calculation**: Next, we calculate: \[ \prod_{k=1}^{3} k = 1 \cdot 2 \cdot 3 = 6 \] Raising this to the power of 4: \[ (6)^{4} = 1296 \] 3. **Final Expression**: Now we combine everything: \[ (7)^{6} + (1296 - 2e^{-x}) \] We know \( 7^6 = 117649 \). Therefore, the final expression becomes: \[ 117649 + 1296 - 2e^{-x} = 118945 - 2e^{-x} \] Thus, the final result is: \[ 118945 - 2e^{-x} \]