To evaluate the limit

$\lim_{x \to -\infty} \left( x e^{\frac{1}{x-2}} + 1 - x \right),$

let's break it into smaller steps and analyze the behavior of each term as $x \to -\infty$.

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Step 1: Simplify the Exponential Term $e^{\frac{1}{x-2}}$

• When $x \to -\infty$, $x - 2 \to -\infty$.
• Therefore, $\frac{1}{x-2} \to 0$.
• Using the property of exponential functions, $e^0 = 1$. Hence, $e^{\frac{1}{x-2}} \to 1$ as $x \to -\infty$.

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Step 2: Substitute $e^{\frac{1}{x-2}}$ in the Expression

Now substitute the approximation $e^{\frac{1}{x-2}} \approx 1$ for large negative $x$: $x e^{\frac{1}{x-2}} + 1 - x \approx x \cdot 1 + 1 - x = x + 1 - x = 1.$

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Step 3: Confirm Behavior of $x e^{\frac{1}{x-2}}$

For completeness:

• $x e^{\frac{1}{x-2}} \approx x \cdot 1 = x$, which cancels with $-x$ in the main expression.
• Any higher-order corrections to $e^{\frac{1}{x-2}}$ are negligible as $x \to -\infty$.

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Final Answer:

$\lim_{x \to -\infty} \left( x e^{\frac{1}{x-2}} + 1 - x \right) = 1.$

Would you like a detailed breakdown of any specific step or have any questions?

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5 Related Questions:

1. How does the behavior of exponential functions affect limits at infinity?
2. What happens to $\frac{1}{x-2}$ when $x \to +\infty$?
3. How do higher-order approximations impact limits in exponential expressions?
4. Can this method be used to evaluate $\lim_{x \to \infty} \left( x e^{\frac{1}{x+3}} - x \right)$?
5. What is the impact of removing or adding constants to terms like $\frac{1}{x-2}$?

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Tip:

Always check the dominant term in an expression when $x$ approaches infinity or negative infinity to simplify the evaluation of limits.