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

$\lim_{x \to \infty} x \sin\left(\frac{1}{x}\right) = 1$

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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.