Prove that limit of ⁴√3x+1/⁴√3x-1 is = 2/⁴√14 as x approaches to 5

We are tasked with proving that:

\[
\lim_{x \to 5} \frac{\sqrt[4]{3x + 1}}{\sqrt[4]{3x - 1}} = \frac{2}{\sqrt[4]{14}}
\]

### Step-by-Step Proof:

1. **Substitute the limit value (x = 5)**:
   First, let's substitute \(x = 5\) into the expression:

   \[
   \frac{\sqrt[4]{3(5) + 1}}{\sqrt[4]{3(5) - 1}} = \frac{\sqrt[4]{15 + 1}}{\sqrt[4]{15 - 1}} = \frac{\sqrt[4]{16}}{\sqrt[4]{14}}
   \]

2. **Simplify the fourth roots**:
   - \(\sqrt[4]{16} = 2\), since \(16 = 2^4\).
   - \(\sqrt[4]{14}\) remains as is because 14 is not a perfect fourth power.

   Thus, the expression becomes:

   \[
   \frac{2}{\sqrt[4]{14}}
   \]

3. **Conclusion**:
   After evaluating the limit at \(x = 5\), we obtain the value:

   \[
   \lim_{x \to 5} \frac{\sqrt[4]{3x + 1}}{\sqrt[4]{3x - 1}} = \frac{2}{\sqrt[4]{14}}
   \]

Therefore, the given limit is indeed equal to \(\frac{2}{\sqrt[4]{14}}\), as required.

### Do you have any further questions or would you like more details?

Here are some related questions to deepen your understanding:

1. What is the general strategy for evaluating limits involving radicals?
2. How does L'Hopital's rule apply to limits involving roots and radicals?
3. Can limits always be evaluated by direct substitution, or are there exceptions?
4. What are some examples where the limit of a rational expression involves indeterminate forms?
5. How do higher-order roots (like cube or fourth roots) affect the behavior of limits?

**Tip**: Always simplify radicals where possible before trying more advanced methods like L'Hopital's rule.

Prove that the limit of ⁴√(3x+1) / ⁴√(3x-1) is equal to 2/⁴√14 as x approaches 5.

This problem demonstrates how to evaluate the limit of a rational expression involving fourth roots as x approaches 5. Learn the step-by-step process of substituting and simplifying radicals to obtain the final result of 2/⁴√14. Suitable for students in Grades 11-12, focusing on limits and radicals.

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