We are tasked to compute the limit:

$\lim_{x \to 6} \frac{\sqrt{x+3} - 3}{x-6}.$

Step-by-Step Solution:

Step 1: Analyze the given expression

As $x \to 6$:

• The numerator becomes $\sqrt{6+3} - 3 = 3 - 3 = 0$,
• The denominator becomes $6 - 6 = 0$.

This results in the indeterminate form $\frac{0}{0}$, so we must simplify further.

Step 2: Rationalize the numerator

Multiply and divide the expression by the conjugate of the numerator $\sqrt{x+3} + 3$: $\frac{\sqrt{x+3} - 3}{x-6} \cdot \frac{\sqrt{x+3} + 3}{\sqrt{x+3} + 3} = \frac{(\sqrt{x+3} - 3)(\sqrt{x+3} + 3)}{(x-6)(\sqrt{x+3} + 3)}.$

Using the difference of squares formula, $(a - b)(a + b) = a^2 - b^2$, the numerator simplifies: $(\sqrt{x+3})^2 - 3^2 = (x+3) - 9 = x - 6.$

Thus, the expression becomes: $\frac{x - 6}{(x - 6)(\sqrt{x+3} + 3)}.$

Step 3: Simplify the fraction

Cancel the common factor $x - 6$ (valid for $x \neq 6$): $\frac{1}{\sqrt{x+3} + 3}.$

Step 4: Evaluate the limit as $x \to 6$

Substitute $x = 6$ into the simplified expression: $\frac{1}{\sqrt{6+3} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}.$

Final Answer:

$\lim_{x \to 6} \frac{\sqrt{x+3} - 3}{x-6} = \boxed{\frac{1}{6}}.$

Would you like additional explanations or related examples?

Related Questions:

1. How can we verify this result using L’Hôpital’s Rule?
2. What happens to the limit if the denominator changes slightly?
3. Why does rationalizing the numerator work for solving this limit?
4. Can this limit be interpreted geometrically?
5. What are other methods to evaluate limits with square roots?

Tip:

When evaluating limits involving square roots, rationalizing the numerator or denominator often simplifies the expression and resolves indeterminate forms.