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We are asked to compute the limit:

\[
\lim_{x \to 1} \frac{\sqrt{3x+1} - \sqrt{5x-1}}{x-1}
\]

### Step 1: Substituting \(x = 1\)
Substitute \(x = 1\) into the expression to check if the limit can be computed directly:
\[
\frac{\sqrt{3(1) + 1} - \sqrt{5(1) - 1}}{1 - 1} = \frac{\sqrt{3 + 1} - \sqrt{5 - 1}}{0} = \frac{\sqrt{4} - \sqrt{4}}{0} = \frac{2 - 2}{0} = \frac{0}{0}
\]

This is an indeterminate form, so we need to apply L'Hopital's Rule or simplify using algebraic techniques.

### Step 2: Applying L'Hopital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), we can apply **L'Hopital's Rule**, which tells us to differentiate the numerator and denominator separately.

The function we need to differentiate is:
\[
f(x) = \frac{\sqrt{3x+1} - \sqrt{5x-1}}{x-1}
\]

#### Derivative of the numerator:
The numerator is \( \sqrt{3x+1} - \sqrt{5x-1} \). Using the chain rule:
\[
\frac{d}{dx} \left( \sqrt{3x+1} \right) = \frac{1}{2\sqrt{3x+1}} \cdot 3 = \frac{3}{2\sqrt{3x+1}}
\]
\[
\frac{d}{dx} \left( \sqrt{5x-1} \right) = \frac{1}{2\sqrt{5x-1}} \cdot 5 = \frac{5}{2\sqrt{5x-1}}
\]

So, the derivative of the numerator is:
\[
\frac{3}{2\sqrt{3x+1}} - \frac{5}{2\sqrt{5x-1}}
\]

#### Derivative of the denominator:
The denominator is \( x - 1 \), and its derivative is just \( 1 \).

### Step 3: Taking the limit
Now we evaluate the limit of the derivative as \( x \to 1 \):
\[
\lim_{x \to 1} \left( \frac{3}{2\sqrt{3x+1}} - \frac{5}{2\sqrt{5x-1}} \right)
\]

Substitute \( x = 1 \):
\[
\frac{3}{2\sqrt{3(1) + 1}} - \frac{5}{2\sqrt{5(1) - 1}} = \frac{3}{2\sqrt{4}} - \frac{5}{2\sqrt{4}} = \frac{3}{2(2)} - \frac{5}{2(2)} = \frac{3}{4} - \frac{5}{4} = \frac{-2}{4} = -\frac{1}{2}
\]

### Final Answer:
The limit is \( \boxed{-\frac{1}{2}} \).

Would you like any additional details or explanations?

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Here are some related questions:
1. How is L'Hopital's Rule applied in general cases of limits?
2. Can the limit be evaluated using algebraic simplification instead of L'Hopital's Rule?
3. What other indeterminate forms exist, besides \( \frac{0}{0} \)?
4. How can the chain rule be applied in differentiating more complex expressions?
5. What is the geometric interpretation of the limit in this problem?

**Tip:** Always check if the direct substitution leads to an indeterminate form before applying any techniques like L'Hopital's Rule.

Limit x->1 ((√(3x+1) - √(5x-1)) / (x-1))

This problem involves calculating the limit of (√(3x+1) - √(5x-1)) / (x-1) as x approaches 1 using L'Hopital's Rule. The problem covers limits, indeterminate forms, and the chain rule for differentiation, suitable for high school or college-level calculus students.

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