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:

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

Let's go through the steps.

### Step 1: Substituting \(x = 5\) into the expression.

First, let's substitute \(x = 5\) directly into the expression and simplify.

\[
\frac{\sqrt[4]{3x+1}}{\sqrt[4]{3x-1}} \Bigg|_{x=5} = \frac{\sqrt[4]{3(5) + 1}}{\sqrt[4]{3(5) - 1}} = \frac{\sqrt[4]{15 + 1}}{\sqrt[4]{15 - 1}} = \frac{\sqrt[4]{16}}{\sqrt[4]{14}}.
\]

### Step 2: Simplifying the expression.

We know that:

\[
\sqrt[4]{16} = 2,
\]

so the expression becomes:

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

This matches the right-hand side of the limit expression we need to prove.

### Conclusion:

Thus, by direct substitution, we have:

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

This confirms that the limit holds true.

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Do you want more details or have any questions?

Here are 5 related questions that might interest you:
1. What is the general strategy for evaluating limits of complex expressions?
2. How do fourth roots behave under limits and substitution?
3. Can L'Hopital's Rule be used in similar limit problems?
4. What are some other techniques for finding limits that involve radical expressions?
5. How would the answer change if the powers were different, say cube roots instead of fourth roots?

**Tip**: When evaluating limits of functions involving roots, always check if substitution works first before considering more advanced techniques.

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

Explore the step-by-step solution to prove that the limit of ⁴√(3x+1)/⁴√(3x-1) is equal to 2/⁴√14 as x approaches 5. This solution involves concepts of limits, radical expressions, and the use of direct substitution. Perfect for students in Grades 11-12 studying advanced calculus.

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