The expression you uploaded is:

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

where $r$ is a constant.

Solution:

To find this limit, let's analyze the behavior of each term inside the parentheses as $x \to \infty$.

1. Dominant Term Analysis:
   • $x$ grows linearly as $x \to \infty$.
   • $e^x$ grows exponentially as $x \to \infty$.
   • $e^{rx}$ grows at the rate of $e^{rx}$, which depends on the value of $r$. If $r > 1$, $e^{rx}$ grows faster than $e^x$, making it the dominant term. If $r < 1$, then $e^x$ is the dominant term.

Since exponential terms dominate over polynomial terms (like $x$), the term $x$ becomes negligible compared to $e^x$ and $e^{rx}$ as $x \to \infty$.

2. Case Analysis Based on $r$:

   • If $r > 1$: $e^{rx}$ will dominate over $e^x$, so $x + e^x + e^{rx} \approx e^{rx}$ as $x \to \infty$.
   • If $r = 1$: $e^x$ and $e^{rx}$ grow at the same rate, so $x + e^x + e^{rx} \approx e^x + e^x = 2e^x$.
   • If $r < 1$: $e^x$ will dominate over $e^{rx}$, so $x + e^x + e^{rx} \approx e^x$.

3. Taking the Limit:

   • For $r > 1$: $\lim_{{x \to \infty}} \left( x + e^x + e^{rx} \right)^{\frac{1}{x}} \approx \lim_{{x \to \infty}} \left( e^{rx} \right)^{\frac{1}{x}} = \lim_{{x \to \infty}} e^{r} = e^r$
   • For $r = 1$: $\lim_{{x \to \infty}} \left( x + e^x + e^{rx} \right)^{\frac{1}{x}} \approx \lim_{{x \to \infty}} \left( 2e^x \right)^{\frac{1}{x}} = \lim_{{x \to \infty}} e = e$
   • For $r < 1$: $\lim_{{x \to \infty}} \left( x + e^x + e^{rx} \right)^{\frac{1}{x}} \approx \lim_{{x \to \infty}} \left( e^x \right)^{\frac{1}{x}} = \lim_{{x \to \infty}} e = e$

Conclusion: