Let's analyze each of the statements carefully, based on the fact that $f$ is continuous on $(-\infty, \infty)$ and that $\lim_{x \to \infty} f(x) = \infty$.

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Statement A: $\lim_{x \to \infty} (f(x) - x)$ exists.

• We know that $\lim_{x \to \infty} f(x) = \infty$, but this gives no information on how fast $f(x)$ grows compared to $x$. The expression $f(x) - x$ could behave in many ways depending on the nature of $f(x)$.
• For example, if $f(x) = x^2$, then $f(x) - x = x^2 - x \to \infty$. In this case, the limit does not exist. However, if $f(x) = x + 1$, then $f(x) - x = 1$, and the limit exists.

Since $f(x)$ can grow faster or slower than $x$, the limit does not necessarily exist for every continuous function satisfying the given conditions.

Conclusion: False.

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Statement B: $\lim_{x \to 0} \frac{1}{f(x)}$ exists.

• This limit depends on the behavior of $f(x)$ as $x \to 0$. There is no information provided about the behavior of $f(x)$ at $x = 0$, only as $x \to \infty$.
• For some functions, $f(x)$ could be zero at $x = 0$ (for example, $f(x) = x^2$), making $\frac{1}{f(x)}$ undefined at $x = 0$. For others, the limit could exist if $f(x)$ behaves well around $x = 0$.

Since we don't have any information about the behavior of $f(x)$ near $0$, we cannot conclude that this limit always exists.

Conclusion: False.

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Statement C: $\lim_{x \to \infty} 4f(x)$ does not exist.

• Given that $\lim_{x \to \infty} f(x) = \infty$, multiplying $f(x)$ by a constant (in this case, 4) will still tend to infinity.
• Thus, $\lim_{x \to \infty} 4f(x) = 4 \cdot \infty = \infty$.

Since the limit goes to infinity, it does not "exist" in a finite sense.

Conclusion: True.

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Statement D: $\lim_{x \to \infty} \frac{4\sin(4x)}{5f(x)}$ exists.

• Since $\lim_{x \to \infty} f(x) = \infty$, as $x \to \infty$, the denominator $5f(x) \to \infty$.
• The numerator $4\sin(4x)$ is bounded because $\sin(4x)$ oscillates between -1 and 1. Therefore, $4\sin(4x)$ oscillates between -4 and 4.
• A bounded numerator divided by an unbounded denominator tends to 0.

Thus, the limit of this expression as $x \to \infty$ is $0$.

Conclusion: True.

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Statement E: $\lim_{x \to 0} 3f(x)$ exists.

• As with Statement B, we do not have any information about the behavior of $f(x)$ as $x \to 0$. The value of $3f(x)$ as $x \to 0$ depends entirely on the behavior of $f(x)$ near 0.
• If $f(x) \to \infty$ or $f(x) \to 0$ as $x \to 0$, the limit may or may not exist.

Since we cannot guarantee the behavior of $f(x)$ near 0, we cannot assert that this limit exists.

Conclusion: False.

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Final Answers:

• A: False
• B: False
• C: True
• D: True
• E: False

Would you like further explanation or more details?

Here are five related questions to deepen your understanding:

1. How would the behavior of $f(x)$ change if we knew $f(x)$ was monotonic?
2. What are some common examples of continuous functions that grow faster than $x$?
3. How does L'Hopital's Rule apply when evaluating limits of oscillatory functions?
4. Can the function $f(x)$ approach infinity more slowly than $x$?
5. What is the behavior of limits involving trigonometric functions over infinity?

Tip: When evaluating limits involving oscillatory functions like sine or cosine, remember that bounded oscillations tend to become negligible when divided by functions that grow without bound.