In this section, we will prove the following limit:
\[ \lim_{x \to \infty} x\sin\left(\frac{1}{x}\right) = 1 \]
We will use two distinct methods for the proof: the Comparison Theorem, exploiting fundamental inequalities of the sine function, and the Taylor series expansion of the function around \( x = 0 \).
Proof using the Comparison Theorem
We use the fundamental inequalities of the sine function:
\[ \sin x \leq x \]
From which it follows that:
\[ \sin\left(\frac{1}{x}\right) \leq \frac{1}{x} \]
Multiplying by \( x \):
\[ x\sin\left(\frac{1}{x}\right) \leq 1 \]
Similarly, we can use the inequality:
\[ \sin x \geq x - \frac{x^3}{6} \]
Which leads to:
\[ x\sin\left(\frac{1}{x}\right) \geq 1 - \frac{1}{6x^2} \]
Applying the Comparison Theorem and taking the limit as \( x \to \infty \), we obtain:
\[ \lim_{x \to \infty} x\sin\left(\frac{1}{x}\right) = 1 \]
Proof using the Taylor Series Expansion
We use the Taylor series expansion of the sine function around zero:
\[ \sin x = x - \frac{x^3}{3!} + \mathcal{O}(x^5) \]
Let \( x = \frac{1}{t} \), so as \( x \to \infty \), we have \( t \to 0 \), and we substitute:
\[ \sin\left(\frac{1}{x}\right) = \frac{1}{x} - \frac{1}{6x^3} + \mathcal{O}\left(\frac{1}{x^5}\right) \]
Multiplying by \( x \):
\[ x\sin\left(\frac{1}{x}\right) = x \left(\frac{1}{x} - \frac{1}{6x^3} + \mathcal{O}\left(\frac{1}{x^5}\right)\right) \]
\[ = 1 - \frac{1}{6x^2} + \mathcal{O}\left(\frac{1}{x^4}\right) \]
Taking the limit as \( x \to \infty \), the term \( \frac{1}{6x^2} \) tends to zero, so:
\[ \lim_{x \to \infty} x\sin\left(\frac{1}{x}\right) = 1 \]