To solve the limit \(\lim_{{x \to \infty}} \left( 5 \cos \frac{2}{x} - 2x \tan \frac{1}{x} \right)\), let's analyze the behavior of each term as \(x\) approaches infinity.

### Step 1: Evaluate \(\cos \frac{2}{x}\)
- As \(x \to \infty\), \(\frac{2}{x} \to 0\).
- The cosine function is continuous, so \(\cos \frac{2}{x} \to \cos 0 = 1\).

Thus, \(\lim_{{x \to \infty}} 5 \cos \frac{2}{x} = 5 \times 1 = 5\).

### Step 2: Evaluate \(\tan \frac{1}{x}\)
- As \(x \to \infty\), \(\frac{1}{x} \to 0\).
- The tangent function is also continuous, so \(\tan \frac{1}{x} \to \tan 0 = 0\).

However, the term \(2x \tan \frac{1}{x}\) involves multiplying \(x\) by \(\tan \frac{1}{x}\), which is small but not zero. We can use the approximation \(\tan y \approx y\) when \(y\) is near 0.

Thus,
\[
2x \tan \frac{1}{x} \approx 2x \cdot \frac{1}{x} = 2
\]
So, \(\lim_{{x \to \infty}} 2x \tan \frac{1}{x} = 2\).

### Step 3: Combine the results
\[
\lim_{{x \to \infty}} \left( 5 \cos \frac{2}{x} - 2x \tan \frac{1}{x} \right) = 5 - 2 = 3
\]

Therefore, the limit is \( \boxed{3} \).

Would you like more details or have any questions? Here are some related questions for further exploration:

1. What is the limit of \( \lim_{{x \to \infty}} x \sin \left( \frac{1}{x} \right) \)?
2. How would you solve \( \lim_{{x \to \infty}} \left( \cos \frac{3}{x} - \tan \frac{2}{x} \right) \)?
3. What happens to \( \lim_{{x \to \infty}} \frac{\sin(1/x)}{1/x} \)?
4. How does L'Hôpital's Rule apply to limits involving trigonometric functions?
5. Can you find the limit \( \lim_{{x \to 0}} x \tan \left( \frac{1}{x} \right) \)?

**Tip:** When dealing with limits involving trigonometric functions as \( x \) approaches infinity, small-angle approximations like \( \sin y \approx y \) and \( \tan y \approx y \) for \( y \) near 0 can be very helpful.

Explore how to evaluate the limit of (5 cos(2/x) - 2x tan(1/x)) as x approaches infinity. Understand the behavior of trigonometric functions and asymptotic analysis in calculus.

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