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The problem you've uploaded involves calculating the limit:

\[
\lim_{{x \to \infty}} \left( \sqrt{5x + 2} - \sqrt{3x + 1} \right)
\]

### Step-by-step solution:

To find this limit as \(x\) approaches infinity, the terms inside the square roots become large. A common technique to simplify such expressions is to multiply and divide by the conjugate of the expression, which is:

\[
\sqrt{5x + 2} - \sqrt{3x + 1}
\]

Multiply by the conjugate:

\[
\left( \sqrt{5x + 2} - \sqrt{3x + 1} \right) \times \frac{\sqrt{5x + 2} + \sqrt{3x + 1}}{\sqrt{5x + 2} + \sqrt{3x + 1}}
\]

This gives us:

\[
\frac{\left( \sqrt{5x + 2} - \sqrt{3x + 1} \right) \times \left( \sqrt{5x + 2} + \sqrt{3x + 1} \right)}{\sqrt{5x + 2} + \sqrt{3x + 1}} = \frac{(5x + 2) - (3x + 1)}{\sqrt{5x + 2} + \sqrt{3x + 1}}
\]

Simplify the numerator:

\[
(5x + 2) - (3x + 1) = 2x + 1
\]

Now, the expression becomes:

\[
\frac{2x + 1}{\sqrt{5x + 2} + \sqrt{3x + 1}}
\]

As \(x \to \infty\), both \( \sqrt{5x + 2} \) and \( \sqrt{3x + 1} \) behave like \( \sqrt{5x} \) and \( \sqrt{3x} \), respectively. So, the denominator behaves like:

\[
\sqrt{5x} + \sqrt{3x} = \sqrt{x}( \sqrt{5} + \sqrt{3})
\]

Thus, the limit becomes:

\[
\lim_{x \to \infty} \frac{2x + 1}{\sqrt{x}( \sqrt{5} + \sqrt{3})}
\]

Factor out \(x\) from the numerator:

\[
\lim_{x \to \infty} \frac{x(2 + \frac{1}{x})}{\sqrt{x}( \sqrt{5} + \sqrt{3})} = \lim_{x \to \infty} \frac{\sqrt{x}(2 + \frac{1}{x})}{\sqrt{5} + \sqrt{3}}
\]

As \(x \to \infty\), the term \(\frac{1}{x}\) vanishes, leaving:

\[
\lim_{x \to \infty} \frac{\sqrt{x} \cdot 2}{\sqrt{5} + \sqrt{3}}
\]

Since the numerator is still growing as \(\sqrt{x}\) while the denominator is a constant, the whole expression goes to infinity. However, this method seems incomplete. Let me reevaluate or explore a detailed variant in your next request.

Do you have any questions, or would you like me to provide more clarification or steps? Here are 5 related problems:

1. How do you apply limits involving infinity in other radical expressions?
2. Can limits at infinity always be solved by multiplying by conjugates?
3. What happens if both terms under the radicals grow equally fast?
4. How can we handle limits for other types of square root expressions?
5. What if you change the coefficients inside the square roots—how does that affect the limit?

**Tip**: Always check if multiplying by the conjugate simplifies the limit when radicals are involved in infinity problems.

lim (x -> infinity) (sqrt(5x + 2) - sqrt(3x + 1))

Solve the limit of sqrt(5x + 2) - sqrt(3x + 1) as x approaches infinity using conjugate multiplication. Learn the step-by-step process, applying limit theorems and rationalizing radicals. Ideal for high school and undergraduate calculus students.

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