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

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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:
$$
118945 - 2e^{-x}
$$
The final result is $$ 118945 - 2e^{-x} $$.

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

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