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) \( \lim _{x \rightarrow \infty} f(x) \)

To evaluate the limits, let's analyze the function step by step for both parts: (a) As \( x \rightarrow -\infty \): We can simplify the expression inside the function \( f(x) \): \[ f(x) = \frac{\sqrt{7 + 9x^3}}{1 - 2x} \] For large negative \( x \), the term \( 9x^3 \) will dominate the 7 in the square root, so we approximate: \[ \sqrt{7 + 9x^3} \sim \sqrt{9x^3} = 3\sqrt{x^3} \] Since \( x \) is negative, \( \sqrt{x^3} = x^{3/2}(-1)^{3/2} = -|x|^{3/2} \). Thus, \[ \sqrt{7 + 9x^3} \sim -3 |x|^{3/2} = -3 (-x)^{3/2} \] Now, for the denominator \( 1 - 2x \), as \( x \) approaches negative infinity, it approaches positive infinity because \( -2x \) will dominate. Putting this together, we have: \[ f(x) \sim \frac{-3(-x)^{3/2}}{1 - 2x} \rightarrow 0 \] Thus, \[ \lim_{x \to -\infty} f(x) = 0 \] (b) As \( x \rightarrow \infty \): We again look at our function: \[ f(x) = \frac{\sqrt{7 + 9x^3}}{1 - 2x} \] For large positive \( x \), similar to part (a), we focus on the term \( 9x^3 \): \[ \sqrt{7 + 9x^3} \sim \sqrt{9x^3} = 3\sqrt{x^3} = 3x^{3/2} \] Now for the denominator \( 1 - 2x \), as \( x \) approaches positive infinity, this goes to negative infinity because \( -2x \) dominates: Putting this all together, we have: \[ f(x) \sim \frac{3x^{3/2}}{1 - 2x} \rightarrow -\frac{3x^{3/2}}{2x} = -\frac{3}{2} x^{1/2} \] As \( x \to \infty \), this expression goes to negative infinity: Thus, \[ \lim_{x \to \infty} f(x) = -\infty \] In summary: (a) \(\lim_{x \rightarrow -\infty} f(x) = 0\) (b) \(\lim_{x \rightarrow \infty} f(x) = -\infty\)