The problem provided is:

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

Step-by-step solution:

1. Term 1: $x$

   As $x \to \infty$, the linear term $x$ obviously tends to $\infty$.

2. Term 2: $\sin(x)$

   The sine function oscillates between -1 and 1 for all $x$. Therefore, as $x \to \infty$, $\sin(x)$ remains bounded, but it doesn't affect the overall limit much because it's relatively small compared to $x$.

3. Term 3: $\sin^2\left( \frac{1}{x} \right)$

   As $x \to \infty$, $\frac{1}{x}$ approaches 0. We know that $\sin(0) = 0$, so:

   $\sin^2\left( \frac{1}{x} \right) \to \sin^2(0) = 0$

   Therefore, this term tends to 0 as $x \to \infty$.

Putting it all together:

$\lim_{{x \to \infty}} \left( x + \sin(x) + \sin^2\left( \frac{1}{x} \right) \right) = \lim_{{x \to \infty}} \left( x + \text{(bounded)} + 0 \right) = \infty$

Thus, the overall limit is:

$\boxed{\infty}$

Would you like further clarification or details on any step?

Related Questions:

1. What is the limit of $x + \sin(x)$ as $x \to \infty$?
2. How does the behavior of $\sin(x)$ change as $x$ increases?
3. What are some common techniques for solving limits involving trigonometric functions?
4. Can oscillating functions like $\sin(x)$ affect the overall growth of functions as $x \to \infty$?
5. What happens if $\sin^2 \left( \frac{1}{x} \right)$ is replaced with $\cos^2 \left( \frac{1}{x} \right)$?

Tip:

When working with limits involving trigonometric functions, it's important to recognize when certain terms are bounded or oscillatory, as they often don't contribute much to the growth of the function at infinity.