We need to evaluate the limit:

$\lim_{x \to \infty} \left( e^x + x \right)^{1/x}.$

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Solution:

1. Factor out $e^x$ from the term $e^x + x$:

$e^x + x = e^x \left( 1 + \frac{x}{e^x} \right).$

This helps us split the expression into two parts.

2. Rewrite the given limit:

$\left( e^x + x \right)^{1/x} = \left[ e^x \left( 1 + \frac{x}{e^x} \right) \right]^{1/x}.$

Using the property of exponents $(ab)^n = a^n \cdot b^n$, we get:

$\left( e^x + x \right)^{1/x} = \left( e^x \right)^{1/x} \cdot \left( 1 + \frac{x}{e^x} \right)^{1/x}.$

3. Simplify each term:

• First term: $\left( e^x \right)^{1/x}$:

$\left( e^x \right)^{1/x} = e^{x \cdot \frac{1}{x}} = e^1 = e.$

• Second term: $\left( 1 + \frac{x}{e^x} \right)^{1/x}$:

As $x \to \infty$, note that $\frac{x}{e^x} \to 0$ because $e^x$ grows much faster than $x$. Therefore:

$1 + \frac{x}{e^x} \to 1.$

Taking the $1/x$-th power:

$\left( 1 + \frac{x}{e^x} \right)^{1/x} \to 1.$

4. Combine the results:

Now combining both terms, we get:

$\left( e^x + x \right)^{1/x} = e \cdot 1 = e.$

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Final Answer:

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

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Would you like a deeper explanation of any step? Let me know!

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Related Questions:

1. What happens if $x \to \infty$ for a similar expression involving $e^{2x}$?
2. How does the rate of growth compare between $e^x$ and polynomial terms like $x^n$?
3. How do we compute limits involving $\log(x)$ or other logarithmic terms?
4. What is the general method for solving exponential and logarithmic limits?
5. How does the dominance of $e^x$ help simplify complicated limits?

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Tip:

When working with exponential and polynomial terms, always factor out the dominant term to simplify the expression.