Let's evaluate the limit:

\[
\lim_{{x \to \infty}} \left( 5 \cos \left( \frac{2}{x} \right) - 2x \tan \left( \frac{1}{x} \right) \right)
\]

### Step 1: Evaluate the first term \( 5 \cos \left( \frac{2}{x} \right) \)
As \( x \) approaches infinity, \(\frac{2}{x}\) approaches 0. Since \(\cos(0) = 1\),

\[
\lim_{{x \to \infty}} 5 \cos \left( \frac{2}{x} \right) = 5 \cdot 1 = 5
\]

### Step 2: Evaluate the second term \( -2x \tan \left( \frac{1}{x} \right) \)
As \( x \) approaches infinity, \(\frac{1}{x}\) approaches 0. We can use the approximation \(\tan(y) \approx y\) for small \( y \). Therefore,

\[
\tan \left( \frac{1}{x} \right) \approx \frac{1}{x}
\]

So, 

\[
-2x \tan \left( \frac{1}{x} \right) \approx -2x \cdot \frac{1}{x} = -2
\]

### Step 3: Combine the results
Now, combine the limits of both terms:

\[
\lim_{{x \to \infty}} \left( 5 \cos \left( \frac{2}{x} \right) - 2x \tan \left( \frac{1}{x} \right) \right) = 5 - 2 = 3
\]

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

Would you like further details or have any questions?

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Here are 5 related questions to consider:

1. How does the limit of trigonometric functions behave as \( x \) approaches infinity?
2. What are common approximations for trigonometric functions near zero?
3. How can you use L'Hôpital's rule to solve limits involving trigonometric functions?
4. What are the limits of functions involving both sine and cosine as \( x \) approaches infinity?
5. How does the limit of a sum of functions differ from the limit of their individual components?

**Tip:** When dealing with trigonometric functions in limits, remember to use small-angle approximations when the angle approaches zero.

Learn how to evaluate the limit lim 5 cos 2/x - 2x tan 1/x as x approaches infinity with a detailed step-by-step solution. This problem covers limits involving trigonometric functions and their small-angle approximations. Suitable for advanced high school students and beyond.

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