Understanding Statements About Real Numbers

The question asks us to determine which of the given statements are true for any real number \(x\). Each statement claims that for a given expression involving an integer \(n\), there exists an integer \(n\) (greater than or equal to a specified value) such that the expression is greater than or equal to \(x\).

For such a statement to be true for any real number \(x\), the sequence defined by the expression must be unbounded above. If a sequence \(a_n\) is unbounded above, it means that for any large number \(M\), there exists an \(n\) such that \(a_n \ge M\). Therefore, for any real number \(x\), we can find an \(n\) such that \(a_n \ge x\).

We need to analyze the behavior of each sequence as \(n\) becomes large (approaches infinity).

Statement 1: \(n^2 \sin\left(\frac{1}{n}\right) \ge x\)

Consider the sequence \(a_n = n^2 \sin\left(\frac{1}{n}\right)\) for \(n \ge 1\).

Let's evaluate the limit of this sequence as \(n \to \infty\). Let \(u = \frac{1}{n}\). As \(n \to \infty\), \(u \to 0^+\). The expression becomes:

\(\lim_{n \to \infty} n^2 \sin\left(\frac{1}{n}\right) = \lim_{u \to 0^+} \left(\frac{1}{u}\right)^2 \sin(u) = \lim_{u \to 0^+} \frac{\sin(u)}{u^2}\)

We can rewrite this as:

\(\lim_{u \to 0^+} \frac{\sin(u)}{u} \cdot \frac{1}{u}\)

We know that \(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\). So the limit becomes:

\(1 \cdot \lim_{u \to 0^+} \frac{1}{u}\)

The limit \(\lim_{u \to 0^+} \frac{1}{u} = +\infty\). Therefore, the sequence \(n^2 \sin\left(\frac{1}{n}\right)\) is unbounded above.

Since the sequence is unbounded above, for any real number \(x\), there exists an integer \(n \ge 1\) such that \(n^2 \sin\left(\frac{1}{n}\right) \ge x\). This statement is true.

Statement 2: \(n \cos\left(\frac{1}{n}\right) \ge x\)

Consider the sequence \(b_n = n \cos\left(\frac{1}{n}\right)\) for \(n \ge 1\).

Let's evaluate the limit of this sequence as \(n \to \infty\). Let \(u = \frac{1}{n}\). As \(n \to \infty\), \(u \to 0^+\). The expression becomes:

\(\lim_{n \to \infty} n \cos\left(\frac{1}{n}\right) = \lim_{u \to 0^+} \frac{1}{u} \cos(u)\)

We can rewrite this as:

\(\lim_{u \to 0^+} \cos(u) \cdot \frac{1}{u}\)

We know that \(\lim_{u \to 0} \cos(u) = \cos(0) = 1\). So the limit becomes:

\(1 \cdot \lim_{u \to 0^+} \frac{1}{u}\)

The limit \(\lim_{u \to 0^+} \frac{1}{u} = +\infty\). Therefore, the sequence \(n \cos\left(\frac{1}{n}\right)\) is unbounded above.

Since the sequence is unbounded above, for any real number \(x\), there exists an integer \(n \ge 1\) such that \(n \cos\left(\frac{1}{n}\right) \ge x\). This statement is true.

Statement 3: \(ne^{-n} \ge x\)

Consider the sequence \(c_n = ne^{-n} = \frac{n}{e^n}\) for \(n \ge 1\).

Let's evaluate the limit of this sequence as \(n \to \infty\). We have an indeterminate form of type \(\frac{\infty}{\infty}\). We can use L'Hopital's rule (treating \(n\) as a real variable):

\(\lim_{n \to \infty} \frac{n}{e^n}\)

Taking the derivative of the numerator and the denominator with respect to \(n\):

\(\lim_{n \to \infty} \frac{\frac{d}{dn}(n)}{\frac{d}{dn}(e^n)} = \lim_{n \to \infty} \frac{1}{e^n}\)

As \(n \to \infty\), \(e^n \to \infty\). So the limit is:

\(\lim_{n \to \infty} \frac{1}{e^n} = 0\)

Since the limit is 0, the sequence \(ne^{-n}\) converges to 0. A convergent sequence is bounded. In this case, for \(n \ge 1\), \(ne^{-n}\) is positive and approaches 0. It is bounded above (e.g., by its value at \(n=1\), which is \(1/e\), or by its maximum value if it occurs at a larger \(n\)). Let M be an upper bound for the sequence. For any \(x > M\), there will be no integer \(n \ge 1\) such that \(ne^{-n} \ge x\).

This statement is false because the sequence is bounded above.

Statement 4: \(n(\log n)^{-1} \ge x\)

Consider the sequence \(d_n = n(\log n)^{-1} = \frac{n}{\log n}\) for \(n \ge 2\) (since \(\log 1 = 0\)).

Let's evaluate the limit of this sequence as \(n \to \infty\). We have an indeterminate form of type \(\frac{\infty}{\infty}\). We can use L'Hopital's rule (treating \(n\) as a real variable):

\(\lim_{n \to \infty} \frac{n}{\log n}\)

Taking the derivative of the numerator and the denominator with respect to \(n\):

\(\lim_{n \to \infty} \frac{\frac{d}{dn}(n)}{\frac{d}{dn}(\log n)} = \lim_{n \to \infty} \frac{1}{1/n} = \lim_{n \to \infty} n\)

The limit is:

\(\lim_{n \to \infty} n = +\infty\)

Therefore, the sequence \(n(\log n)^{-1}\) is unbounded above.

Since the sequence is unbounded above, for any real number \(x\), there exists an integer \(n \ge 2\) such that \(n(\log n)^{-1} \ge x\). This statement is true.

Summary of Statements

Based on the analysis:

• Statement 1 is true because the sequence \(n^2 \sin\left(\frac{1}{n}\right)\) is unbounded above.
• Statement 2 is true because the sequence \(n \cos\left(\frac{1}{n}\right)\) is unbounded above.
• Statement 3 is false because the sequence \(ne^{-n}\) is bounded above.
• Statement 4 is true because the sequence \(n(\log n)^{-1}\) is unbounded above.