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

To prove that the limit of \(\frac{\sqrt[4]{3x+1}}{\sqrt[4]{3x-1}}\) as \(x\) approaches 5 equals \(\frac{2}{\sqrt[4]{14}}\), we will use the **epsilon-delta definition of a limit**. Specifically, we want to prove that for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that:

\[
\left| \frac{\sqrt[4]{3x+1}}{\sqrt[4]{3x-1}} - \frac{2}{\sqrt[4]{14}} \right| < \epsilon \quad \text{whenever} \quad 0 < |x - 5| < \delta.
\]

### Step 1: Identify the target limit
Let \( L = \frac{2}{\sqrt[4]{14}} \). This is the value we expect the function to approach as \(x\) approaches 5.

We begin by computing the value of the function at \(x = 5\):
\[
f(x) = \frac{\sqrt[4]{3x+1}}{\sqrt[4]{3x-1}}.
\]
At \(x = 5\):
\[
f(5) = \frac{\sqrt[4]{3(5)+1}}{\sqrt[4]{3(5)-1}} = \frac{\sqrt[4]{16}}{\sqrt[4]{14}} = \frac{2}{\sqrt[4]{14}} = L.
\]

So we are proving:
\[
\lim_{x \to 5} \frac{\sqrt[4]{3x+1

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

Learn how to prove the limit of the function ⁴√(3x+1) / ⁴√(3x-1) as x approaches 5 using the epsilon-delta definition. This detailed step-by-step guide walks through key concepts in limits, radical expressions, and the formal limit proof technique. Suitable for university-level calculus students.

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