The expression in the image you uploaded is:

$\lim_{x \to \infty} \left( 1 + \sin\left(\frac{1}{x}\right) \right)^{5x}$

To solve this, we first need to analyze the behavior of the components as $x \to \infty$:

1. As $x \to \infty$, $\frac{1}{x}$ approaches zero. Therefore, $\sin\left(\frac{1}{x}\right)$ also approaches zero. So, the base $1 + \sin\left(\frac{1}{x}\right)$ approaches 1.

2. The exponent $5x$ grows without bound as $x \to \infty$.

We have a limit of the form $\lim_{x \to \infty} \left( 1 + \text{something small} \right)^{\text{large number}}$. This suggests we can apply a standard limit result often associated with the exponential function.

We can approximate $\sin\left(\frac{1}{x}\right) \approx \frac{1}{x}$ for large $x$, leading to:

$\lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^{5x}$

This is a well-known limit that results in $e^5$, because:

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

So, applying the exponent 5:

$\lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^{5x} = e^5$

Thus, the value of the limit is $e^5$.

Would you like a more detailed explanation or have any questions?

Here are 5 related questions to explore further:

1. What is the exponential limit rule, and how does it work in calculus?
2. How do small angle approximations, like $\sin(x) \approx x$, help simplify limits?
3. What is the behavior of other trigonometric functions as $x \to \infty$?
4. How does this method apply to limits involving logarithms?
5. What are the applications of limits of this type in real-world problems?

Tip: When solving limits with expressions that approach 1 or 0 as $x \to \infty$, it's often helpful to apply approximations or known limit results.