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 1: Evaluate the expression at \( x = 5 \)

We begin by substituting \( x = 5 \) directly into the expression:

1. **Evaluate \( 3x + 1 \) when \( x = 5 \):**
   \[
   3(5) + 1 = 15 + 1 = 16
   \]
   So, \( \sqrt[4]{3x + 1} = \sqrt[4]{16} = 2 \).

2. **Evaluate \( 3x - 1 \) when \( x = 5 \):**
   \[
   3(5) - 1 = 15 - 1 = 14
   \]
   So, \( \sqrt[4]{3x - 1} = \sqrt[4]{14} \).

Thus, when \( x = 5 \), the original expression becomes:
\[
\frac{\sqrt[4]{16}}{\sqrt[4]{14}} = \frac{2}{\sqrt[4]{14}}.
\]

### Step 2: Conclusion

Since the function is continuous and substituting \( x = 5 \) into the expression gives us exactly the value on the right-hand side of the equation, we can conclude that:
\[
\lim_{x \to 5} \frac{\sqrt[4]{3x+1}}{\sqrt[4]{3x-1}} = \frac{2}{\sqrt[4]{14}}.
\]

Thus, the limit is proven to be \( \frac{2}{\sqrt[4]{14}} \).

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Do you have any questions or would you like further details on any of the steps?

Here are 5 related questions to expand on this topic:
1. How would the limit change if the power was different, say a cube root?
2. What happens to the limit as \( x \) approaches a different value, like \( x = 0 \)?
3. How can we apply L'Hôpital's Rule to limits of similar expressions?
4. What are the conditions for a function to be continuous and why is it important for evaluating limits?
5. How can we evaluate limits that result in indeterminate forms like \( \frac{0}{0} \)?

**Tip:** When evaluating limits, always check for continuity of the function at the point you're approaching. This often allows you to directly substitute the value, simplifying the process.

Explore how to prove the limit of the function ⁴√(3x+1)/⁴√(3x-1) as x approaches 5, using concepts of limits and continuity. Learn step-by-step how to evaluate the limit and understand key radical function properties. Suitable for Grades 11-12.

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