We need to evaluate the limit of the expression as $x \to \infty$:

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

Let's break this down step by step:

Step 1: Analyze small-angle approximations

As $x \to \infty$, the terms $\frac{1}{x}$ and $\frac{2}{x}$ become small, so we can apply small-angle approximations for trigonometric functions:

• For $\tan\left(\frac{1}{x}\right)$, use $\tan z \approx z$ for small $z$, so: $\tan\left(\frac{1}{x}\right) \approx \frac{1}{x}.$

• For $\sin\left(\frac{1}{x}\right)$, use $\sin z \approx z$ for small $z$, so: $\sin\left(\frac{1}{x}\right) \approx \frac{1}{x}.$

• For $\cos\left(\frac{2}{x}\right)$, use $\cos z \approx 1$ for small $z$, so: $\cos\left(\frac{2}{x}\right) \approx 1.$

Step 2: Simplify the expression

Using the approximations, the numerator becomes:

$2x^2 \cdot \frac{1}{x} - x \cdot \frac{1}{x} + \frac{1}{x} = 2x - 1 + \frac{1}{x}.$

The denominator becomes:

$x \cdot 1 = x.$

So, the expression simplifies to:

$\frac{2x - 1 + \frac{1}{x}}{x}.$

Step 3: Evaluate the limit as $x \to \infty$

As $x \to \infty$, the terms $\frac{1}{x}$ approach zero. Therefore, the expression becomes:

$\frac{2x - 1}{x} = 2 - \frac{1}{x}.$

Again, as $x \to \infty$, $\frac{1}{x} \to 0$, so the final result is:

$2.$

Final Answer:

The value of the limit is $\boxed{2}$.

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Do you want any further details or have any questions?

Here are five related questions to deepen your understanding:

1. What is the limit of $\frac{\sin(1/x)}{x}$ as $x \to \infty$?
2. How would the result change if the numerator had $\tan\left(\frac{2}{x}\right)$ instead of $\tan\left(\frac{1}{x}\right)$?
3. How does the small-angle approximation work for higher powers of $\frac{1}{x}$?
4. What would the limit be if $x \to 0$ instead of $x \to \infty$ in the given expression?
5. How do you apply L'Hopital's rule to limits involving trigonometric functions?

Tip: When solving limits involving trigonometric functions, small-angle approximations for $\sin z$, $\cos z$, and $\tan z$ are very helpful when $z$ approaches 0.