lim
𝑥 →1
√2 − 𝑥 − 1
2 − √𝑥 + 3

To solve the limit

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

we can approach this by rationalizing the numerator and the denominator.

### Step 1: Rationalize the Numerator
The numerator is \( \sqrt{2 - x} - 1 \). Multiply both the numerator and denominator by the conjugate of the numerator:

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

Simplifying the numerator:

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

### Step 2: Rationalize the Denominator
Now, we rationalize the denominator \(2 - \sqrt{x + 3}\). Multiply both the numerator and denominator by the conjugate of the denominator:

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

Simplify the denominator:

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

### Step 3: Take the Limit
As \( x \) approaches 1, the term \( 1 - x \) tends to 0. Cancel the \( 1 - x \) in the numerator and denominator:

\[
\lim_{x \to 1} \frac{2 + \sqrt{x + 3}}{(\sqrt{2 - x} + 1)} = \frac{2 + \sqrt{1 + 3}}{(\sqrt{2 - 1} + 1)} = \frac{2 + 2}{1 + 1} = \frac{4}{2} = 2.
\]

So, the limit is \(\boxed{2}\).

Do you want details or have any questions? Here are some related questions to explore:

1. How does rationalizing help in solving limits?
2. What are conjugates, and why are they useful in limit problems?
3. Can you find a limit where direct substitution fails but rationalizing works?
4. How would this limit change if the square roots were replaced with cube roots?
5. What happens if we reverse the operations (denominator becomes numerator)?

**Tip:** When faced with indeterminate forms like \( \frac{0}{0} \), rationalizing can often help simplify the expression.

Learn how to evaluate the limit lim_{x \to 1} (sqrt(2 - x) - 1) / (2 - sqrt(x + 3)) with a step-by-step solution. Explore rationalization techniques and the simplification of square roots in limit problems. Suitable for advanced high school students and above.

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