\( \operatorname { Lim } _ { x \rightarrow \infty } \frac { \sqrt[ 3 ] { 8 \sqrt { x ^ { 5 } + 2 x } + 4 x } + 3 \sqrt[ 3 ] { x ^ { 2 } } \sqrt[ 6 ] { x } } { \sqrt { 2 + \sqrt { 3 + x \sqrt { x ^ { 3 } } } } } \)

To solve the limit \( \operatorname{Lim}_{x \rightarrow \infty} \frac{\sqrt[3]{8\sqrt{x^5 + 2x} + 4x} + 3\sqrt[3]{x^2}\sqrt[6]{x}}{\sqrt{2 + \sqrt{3 + x\sqrt{x^3}}}} \), we'll analyze the behaviors of the numerator and denominator as \( x \) approaches infinity. First, for the **numerator**: 1. Inside the cube root, as \( x \rightarrow \infty \), \( x^5 \) dominates \( 2x \): \[ \sqrt{x^5 + 2x} \sim \sqrt{x^5} = x^{5/2} \] Thus, \[ 8\sqrt{x^5 + 2x} + 4x \sim 8x^{5/2} \] Therefore, \[ \sqrt[3]{8\sqrt{x^5 + 2x} + 4x} \sim \sqrt[3]{8x^{5/2}} = 2x^{5/6} \] 2. For \( 3\sqrt[3]{x^2}\sqrt[6]{x} \): \[ \sqrt[3]{x^2}\sqrt[6]{x} = x^{2/3 + 1/6} = x^{4/6 + 1/6} = x^{5/6} \] So, \[ 3\sqrt[3]{x^2}\sqrt[6]{x} = 3x^{5/6} \] Combining these, we get the numerator: \[ \sqrt[3]{8\sqrt{x^5 + 2x} + 4x} + 3\sqrt[3]{x^2}\sqrt[6]{x} \sim 2x^{5/6} + 3x^{5/6} = 5x^{5/6} \] Next, consider the **denominator**: 1. Inside the square root, \[ x\sqrt{x^3} = x^{1/2}x^{3/2} = x^2 \] Thus, \( 3 + x\sqrt{x^3} \sim 3 + x^2 \) as \( x \) becomes very large. This means the term approximates: \[ \sqrt{2 + \sqrt{3 + x\sqrt{x^3}}} \sim \sqrt{2 + \sqrt{x^2}} = \sqrt{2 + x} \sim \sqrt{x} \text{ (for large } x\text{)} \] So the denominator becomes: \[ \sqrt{2 + \sqrt{3 + x\sqrt{x^3}}} \sim \sqrt{x} \] Putting this all together, we have: \[ \frac{5x^{5/6}}{\sqrt{x}} = \frac{5x^{5/6}}{x^{1/2}} = 5x^{5/6 - 3/6} = 5x^{2/6} = 5x^{1/3} \] As \( x \rightarrow \infty \), \( 5x^{1/3} \rightarrow \infty \). Therefore, the final result of the limit is: \[ \operatorname{Lim}_{x \rightarrow \infty} \frac{\sqrt[3]{8\sqrt{x^5 + 2x} + 4x} + 3\sqrt[3]{x^2}\sqrt[6]{x}}{\sqrt{2 + \sqrt{3 + x\sqrt{x^3}}}} = \infty \]