Given the function: \[ f(x)=\frac{\sqrt{7+9 x^{3}}}{1-2 x} \] evaluate the following limits: (a) \( \lim _{x \rightarrow-\infty} f(x) \) (b) \( \lim _{x \rightarrow \infty} f(x) \)

We begin with the function   f(x) = (√(7 + 9x³)) / (1 – 2x). Before evaluating the limits, note that for f(x) to be a real–valued function the radicand must be nonnegative:   7 + 9x³ ≥ 0 ⟹ x³ ≥ –7/9 ⟹ x ≥ ∛(–7/9). Thus, the domain of f is x ≥ ∛(–7/9) (approximately x ≥ –0.92). With this in mind, let’s address the two parts. ────────────────────────────── (a) Limit as x → –∞ Since the domain of f is limited to x values greater than or equal to ∛(–7/9), there is no “real” value of x less than ∛(–7/9); in other words, the function is not defined (in the real numbers) for x near –∞. Hence, one cannot take the limit as x → –∞ in the usual real–analysis sense. Answer for (a): The limit as x → –∞ does not exist in the real numbers (since f(x) is only defined for x ≥ ∛(–7/9)). ────────────────────────────── (b) Limit as x → +∞ For large positive x, we can analyze the behavior of the numerator and denominator separately. 1. The numerator is √(7 + 9x³). As x → +∞, the term 9x³ dominates the constant 7. Thus,   7 + 9x³ ~ 9x³,  and therefore,   √(7 + 9x³) ~ √(9x³) = 3x^(3/2). 2. The denominator is 1 – 2x, and for large x, the –2x term dominates so that   1 – 2x ~ –2x. Thus, for large x, we have approximately:   f(x) ~ (3x^(3/2)) / (–2x) = –(3/2) x^(3/2 – 1) = –(3/2) x^(1/2). Since x^(1/2) grows without bound as x → +∞, it follows that   –(3/2) x^(1/2) → –∞. Answer for (b): limₓ→∞ f(x) = –∞. ────────────────────────────── Summary of Answers: (a) The limit as x → –∞ does not exist (within the real numbers) because the function is only defined for x ≥ ∛(–7/9). (b) limₓ→∞ f(x) = –∞.