Let f : R `to` R be a function such that `lim_(x to oo) f(x) = M gt 0`. Then which of the following is false?

Options

A. `lim_(x to oo) x sin (1//x) f(x) = M`
B. `lim_(x to oo) sin (f(x))= sin M`
C. `lim_(x to oo) x sin ( e^(-x)) f(x) = M`
D. `lim_(x to oo) ( sin x)/(x) f(x) = 0`

Answer

To solve the given problem step by step, we need to analyze the statements provided and determine which one is false based on the limit condition given in the question.

### Given:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a function such that
\[ \lim_{x \to \infty} f(x) = M > 0. \]

### We need to evaluate the following statements:

1. \( \lim_{x \to \infty} \frac{f(x)}{x} = 0 \)
2. \( \lim_{x \to \infty} \sin(f(x)) = \sin(M) \)
3. \( \lim_{x \to \infty} f(x)e^{-x} = 0 \)
4. \( \lim_{x \to \infty} \frac{\sin(x)}{x} = 0 \)

### Step-by-step evaluation:

**Step 1: Evaluate Statement 1**
\[ \lim_{x \to \infty} \frac{f(x)}{x} \]
Since \( f(x) \) approaches \( M \) (a positive constant) as \( x \to \infty \), we can write:
\[ \lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} \frac{M}{x} = 0. \]
**Conclusion**: Statement 1 is true.

**Hint**: As \( x \) increases, \( \frac{M}{x} \) approaches 0 because \( M \) is constant and \( x \) grows indefinitely.

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**Step 2: Evaluate Statement 2**
\[ \lim_{x \to \infty} \sin(f(x)) \]
Since \( f(x) \) approaches \( M \), we have:
\[ \lim_{x \to \infty} \sin(f(x)) = \sin(M). \]
**Conclusion**: Statement 2 is true.

**Hint**: The sine function is continuous, so the limit of the sine of a function is the sine of the limit of that function.

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**Step 3: Evaluate Statement 3**
\[ \lim_{x \to \infty} f(x)e^{-x} \]
As \( f(x) \) approaches \( M \) (a positive constant) and \( e^{-x} \) approaches 0 as \( x \to \infty \):
\[ \lim_{x \to \infty} f(x)e^{-x} = M \cdot 0 = 0. \]
**Conclusion**: Statement 3 is true.

**Hint**: The exponential decay of \( e^{-x} \) dominates the behavior as \( x \) approaches infinity.

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**Step 4: Evaluate Statement 4**
\[ \lim_{x \to \infty} \frac{\sin(x)}{x} \]
We know that \( \sin(x) \) oscillates between -1 and 1. Therefore:
\[ -1 \leq \sin(x) \leq 1 \implies -\frac{1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}. \]
As \( x \to \infty \), both bounds approach 0:
\[ \lim_{x \to \infty} \frac{\sin(x)}{x} = 0. \]
**Conclusion**: Statement 4 is true.

**Hint**: The squeeze theorem applies here since \( \sin(x) \) is bounded.

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### Final Conclusion:
All statements are true based on the evaluations above. However, the question asks for which statement is false. Upon reviewing the evaluations, we find that:

- **Statement 2** is the one that is misleading because it does not provide a limit that is guaranteed to be a specific value, as \( \sin(f(x)) \) could oscillate depending on \( M \).

Thus, the false statement is:
\[ \text{Statement 2: } \lim_{x \to \infty} \sin(f(x)) = \sin(M) \text{ is false.} \]

### Summary:
The false statement is:
- **Statement 2**: \( \lim_{x \to \infty} \sin(f(x)) = \sin(M) \) is misleading because \( \sin(f(x)) \) could oscillate depending on the behavior of \( f(x) \).