lim x$$ >\infty\text{  xsin}\frac{1}{x} $$

We need to evaluate the limit:  

\[
\lim_{x \to \infty} x \sin\left(\frac{1}{x}\right)
\]

### Step 1: Approximate $\sin(1/x)$  
For small values of $y$, we use the first-order approximation:  

\[
\sin y \approx y \quad \text{as } y \to 0
\]

Substituting $y = \frac{1}{x}$, we get:  

\[
\sin\left(\frac{1}{x}\right) \approx \frac{1}{x} \quad \text{for large } x
\]

### Step 2: Multiply by $x$  

\[
x \sin\left(\frac{1}{x}\right) \approx x \cdot \frac{1}{x} = 1
\]

### Conclusion  
\[
\lim_{x \to \infty} x \sin\left(\frac{1}{x}\right) = 1
\]

Would you like a more detailed explanation?  

### Related Questions  
1. What is the Taylor series expansion of $\sin x$ and how is it used in limits?  
2. How does L'Hôpital’s Rule apply to similar limit problems?  
3. What happens if we replace $\sin(1/x)$ with $\tan(1/x)$?  
4. How does this limit change if we consider $x^2 \sin(1/x)$ instead?  
5. Can we use the squeeze theorem to confirm this result?  

#### Tip  
For small angles ($x \to 0$), $\sin x \approx x$ is a useful approximation in limit problems.

Learn how to evaluate the limit lim x → ∞ x sin(1/x) using approximations and limit properties. This problem involves small angle approximations and is suitable for students in Grades 11-12.

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