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

Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow -\infty}\left(\frac{\sqrt{7+9x^{3}}}{1-2x}\right)\) - step1: Evaluate the limit: \(\textrm{Undefined}\) Calculate the limit \( \lim_{x\rightarrow \infty} \frac{\sqrt{7+9 x^{3}}}{1-2 x} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow +\infty}\left(\frac{\sqrt{7+9x^{3}}}{1-2x}\right)\) - step1: Rewrite the expression: \(\lim _{x\rightarrow +\infty}\left(\frac{\sqrt{\frac{7}{x^{3}}+9}\times \left|x\right|\times \sqrt{x}}{\left(\frac{1}{x}-2\right)x}\right)\) - step2: Simplify: \(\lim _{x\rightarrow +\infty}\left(\frac{x^{\frac{1}{2}}\sqrt{\frac{7}{x^{3}}+9}}{\frac{1}{x}-2}\right)\) - step3: Rewrite the expression: \(\frac{\lim _{x\rightarrow +\infty}\left(x^{\frac{1}{2}}\sqrt{\frac{7}{x^{3}}+9}\right)}{\lim _{x\rightarrow +\infty}\left(\frac{1}{x}-2\right)}\) - step4: Calculate: \(\frac{+\infty}{\lim _{x\rightarrow +\infty}\left(\frac{1}{x}-2\right)}\) - step5: Calculate: \(\frac{+\infty}{-2}\) - step6: Divide the terms: \(-\infty\) Let's evaluate the limits step by step. ### (a) \( \lim _{x \rightarrow -\infty} f(x) \) The function is given by: \[ f(x) = \frac{\sqrt{7 + 9x^3}}{1 - 2x} \] As \( x \) approaches \(-\infty\): - The term \( 9x^3 \) dominates the expression inside the square root, so we can approximate: \[ \sqrt{7 + 9x^3} \approx \sqrt{9x^3} = 3\sqrt{x^3} = 3x^{3/2} \] However, since \( x \) is negative, \( \sqrt{x^3} \) will be negative, leading to: \[ \sqrt{7 + 9x^3} \approx -3(-x)^{3/2} = -3(-1)^{3/2}(-x)^{3/2} = -3(-x)^{3/2} \] - The denominator \( 1 - 2x \) approaches \( +\infty \) as \( x \) approaches \(-\infty\). Thus, we have: \[ f(x) \approx \frac{-3(-x)^{3/2}}{+\infty} \rightarrow 0 \] However, the limit is actually undefined because the square root of a negative number is not defined in the real number system. ### Conclusion for (a): \[ \lim _{x \rightarrow -\infty} f(x) = \text{Undefined} \] ### (b) \( \lim _{x \rightarrow \infty} f(x) \) As \( x \) approaches \( \infty \): - The term \( 9x^3 \) again dominates the expression inside the square root: \[ \sqrt{7 + 9x^3} \approx \sqrt{9x^3} = 3\sqrt{x^3} = 3x^{3/2} \] - The denominator \( 1 - 2x \) approaches \( -\infty \) as \( x \) approaches \( \infty\). Thus, we have: \[ f(x) \approx \frac{3x^{3/2}}{-\infty} \rightarrow -\infty \] ### Conclusion for (b): \[ \lim _{x \rightarrow \infty} f(x) = -\infty \] ### Final Results: - (a) \( \lim _{x \rightarrow -\infty} f(x) = \text{Undefined} \) - (b) \( \lim _{x \rightarrow \infty} f(x) = -\infty \)