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

We are tasked with proving that:

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

using the **epsilon-delta** definition of a limit.

### **Step 1: Recall the epsilon-delta definition**

The epsilon-delta definition of a limit states that:
\[
\lim_{x \to a} f(x) = L \quad \text{means} \quad \forall \epsilon > 0, \, \exists \delta > 0 \quad \text{such that} \quad 0 < |x - a| < \delta \implies |f(x) - L| < \epsilon
\]

For this particular problem, we want to show that for any \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that:
\[
0 < |x - 5| < \delta \implies \left| \frac{\sqrt[4]{3x + 1}}{\sqrt[4]{3x - 1}} - \frac{2}{\sqrt[4]{14}} \right| < \epsilon
\]

### **Step 2: Simplify and find the limit value**

We first compute the limit of the given expression as \( x \to 5 \).

- When \( x = 5 \), the expression becomes:
\[
\frac{\sqrt[4]{3(5) + 1}}{\sqrt[4]{3(5) - 1}} = \frac{\sqrt[4]{16}}{\sqrt[4]{14}} = \frac{2}{\sqrt[4]{14}}
\]
This confirms that the limit is indeed \( \frac{2}{\sqrt[4]{14}} \).

### **Step 3: Set up the inequality**

We now need to show that for any \( \epsilon > 0 \), we can find a corresponding \( \delta \) such that:

\[
\left| \frac{\sqrt[4]{3x + 1}}{\sqrt[4]{3x - 1}} - \frac{2}{\sqrt[4]{14}} \right| < \epsilon \quad \text{whenever}

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

This page provides a detailed epsilon-delta proof for the limit of ⁴√(3x+1)/⁴√(3x-1) as x approaches 5, showing that it equals 2/⁴√14. Understand key concepts in calculus limits and the epsilon-delta definition of limits with a step-by-step guide.

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