Math Problem Statement
jawab soal diatas
Solution
The given problem asks to find the limit:
This is a typical limit problem involving a difference of square roots. Let's go step by step to solve it:
Step 1: Check direct substitution
First, substitute directly into the expression:
= \frac{\sqrt{18 - 2} - \sqrt{9 + 7}}{0} = \frac{\sqrt{16} - \sqrt{16}}{0} = \frac{4 - 4}{0} = \frac{0}{0}$$ This results in an indeterminate form $$\frac{0}{0}$$, so we need to simplify using other techniques, such as multiplying by the conjugate. ### Step 2: Multiply by the conjugate To resolve the indeterminate form, multiply both the numerator and the denominator by the conjugate of the numerator: $$\frac{\sqrt{6x - 2} - \sqrt{3x + 7}}{x - 3} \cdot \frac{\sqrt{6x - 2} + \sqrt{3x + 7}}{\sqrt{6x - 2} + \sqrt{3x + 7}}$$ This will simplify the numerator using the difference of squares: $$= \frac{(6x - 2) - (3x + 7)}{(x - 3)(\sqrt{6x - 2} + \sqrt{3x + 7})} = \frac{6x - 2 - 3x - 7}{(x - 3)(\sqrt{6x - 2} + \sqrt{3x + 7})} = \frac{3x - 9}{(x - 3)(\sqrt{6x - 2} + \sqrt{3x + 7})}$$ ### Step 3: Factor and cancel terms The numerator $$3x - 9$$ can be factored: $$= \frac{3(x - 3)}{(x - 3)(\sqrt{6x - 2} + \sqrt{3x + 7})}$$ Now, cancel the $$x - 3$$ term in both the numerator and denominator: $$= \frac{3}{\sqrt{6x - 2} + \sqrt{3x + 7}}$$ ### Step 4: Substitute $$x = 3$$ into the simplified expression Now, substitute $$x = 3$$ into the simplified expression: $$= \frac{3}{\sqrt{6(3) - 2} + \sqrt{3(3) + 7}} = \frac{3}{\sqrt{18 - 2} + \sqrt{9 + 7}} = \frac{3}{\sqrt{16} + \sqrt{16}} = \frac{3}{4 + 4} = \frac{3}{8}$$ Thus, the value of the limit is: $$\boxed{\frac{3}{8}}$$ The correct answer is **C**. --- Do you have any further questions, or would you like more details on any step? Here are some related questions you could explore: 1. How can limits involving square roots be simplified in general? 2. What is the difference of squares and how is it used in limit problems? 3. What happens if a limit does not yield an indeterminate form like $$\frac{0}{0}$$? 4. Can L'Hôpital's Rule be applied in this case? Why or why not? 5. How do you handle limits at infinity involving square roots? **Tip:** Always check for indeterminate forms like $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$ before applying simplification techniques like conjugates or factoring.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Square Roots
Indeterminate Forms
Conjugates
Formulas
Difference of squares: a^2 - b^2 = (a - b)(a + b)
Basic limit properties
Theorems
Limit Laws
Difference of Squares Theorem
Suitable Grade Level
Grades 11-12
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