https://file.aiquickdraw.com/mathbot/file/1726045588489sp2x7fdv.jpg

The expression given in the image is the limit:

\[
\lim_{x \to 1} \frac{\sqrt{3x + 1} - \sqrt{2x + 2}}{\sqrt{8x + 1} - 3}
\]

We can attempt to simplify and solve this limit. Here's the approach:

### Step 1: Plug in \( x = 1 \) to check if the limit is indeterminate
Substituting \( x = 1 \):

\[
\frac{\sqrt{3(1) + 1} - \sqrt{2(1) + 2}}{\sqrt{8(1) + 1} - 3} = \frac{\sqrt{3 + 1} - \sqrt{2 + 2}}{\sqrt{8 + 1} - 3} = \frac{\sqrt{4} - \sqrt{4}}{\sqrt{9} - 3} = \frac{2 - 2}{3 - 3} = \frac{0}{0}
\]

Since the result is \( \frac{0}{0} \), we have an indeterminate form, so we need to simplify the expression using algebraic techniques like multiplying by the conjugate.

### Step 2: Multiply numerator and denominator by their respective conjugates
We first multiply the numerator and the denominator by their conjugates to eliminate the square roots:

#### Numerator:
\[
\text{Conjugate of } \sqrt{3x + 1} - \sqrt{2x + 2} \text{ is } \sqrt{3x + 1} + \sqrt{2x + 2}
\]
Multiply the numerator by \( \frac{\sqrt{3x + 1} + \sqrt{2x + 2}}{\sqrt{3x + 1} + \sqrt{2x + 2}} \):

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

#### Denominator:
\[
\text{Conjugate of } \sqrt{8x + 1} - 3 \text{ is } \sqrt{8x + 1} + 3
\]
Multiply the denominator by \( \frac{\sqrt{8x + 1} + 3}{\sqrt{8x + 1} + 3} \):

\[
\frac{[\sqrt{8x + 1} - 3][\sqrt{8x + 1} + 3]}{\sqrt{8x + 1} + 3} = \frac{(8x + 1) - 9}{\sqrt{8x + 1} + 3} = \frac{8x + 1 - 9}{\sqrt{8x + 1} + 3} = \frac{8x - 8}{\sqrt{8x + 1} + 3} = \frac{8(x - 1)}{\sqrt{8x + 1} + 3}
\]

### Step 3: Simplified limit expression
Now, we rewrite the entire expression:

\[
\lim_{x \to 1} \frac{\frac{x - 1}{\sqrt{3x + 1} + \sqrt{2x + 2}}}{\frac{8(x - 1)}{\sqrt{8x + 1} + 3}} = \lim_{x \to 1} \frac{x - 1}{8(x - 1)} \cdot \frac{\sqrt{8x + 1} + 3}{\sqrt{3x + 1} + \sqrt{2x + 2}}
\]

Cancel out the \( x - 1 \) terms:

\[
= \lim_{x \to 1} \frac{1}{8} \cdot \frac{\sqrt{8x + 1} + 3}{\sqrt{3x + 1} + \sqrt{2x + 2}}
\]

### Step 4: Substitute \( x = 1 \) again
Now, substitute \( x = 1 \):

\[
= \frac{1}{8} \cdot \frac{\sqrt{8(1) + 1} + 3}{\sqrt{3(1) + 1} + \sqrt{2(1) + 2}} = \frac{1}{8} \cdot \frac{\sqrt{9} + 3}{\sqrt{4} + \sqrt{4}} = \frac{1}{8} \cdot \frac{3 + 3}{2 + 2} = \frac{1}{8} \cdot \frac{6}{4} = \frac{1}{8} \cdot \frac{3}{2} = \frac{3}{16}
\]

### Final Answer:
\[
\lim_{x \to 1} \frac{\sqrt{3x + 1} - \sqrt{2x + 2}}{\sqrt{8x + 1} - 3} = \frac{3}{16}
\]

Let me know if you'd like further details or have any questions! Here are five related questions:

1. What is the general strategy for solving limits involving square roots?
2. How do conjugates help in simplifying limits?
3. What are some alternative methods for solving indeterminate forms in limits?
4. How do you apply L'Hopital's rule to limits that yield indeterminate forms?
5. What are common techniques to simplify limits involving rational functions?

**Tip:** When dealing with square roots in limits, multiplying by the conjugate is often an effective way to simplify the expression.

lim(x → 1) [(√(3x + 1) - √(2x + 2)) / (√(8x + 1) - 3)]

This detailed guide explains how to solve the limit lim(x → 1) [(√(3x + 1) - √(2x + 2)) / (√(8x + 1) - 3)] using algebraic techniques, including multiplying by conjugates. Learn how to approach indeterminate forms with step-by-step solutions.

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