| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve the limit \(\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}\), we can apply a technique known as rationalization or multiplying by the conjugate to remove the radical from the denominator. The idea is to multiply the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{x} + 3\). \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \] By doing so, we get: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} \] The denominator simplifies to a difference of squares: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} \] Since \(x \neq 9\) in the limit, we can safely cancel out the \(x - 9\) terms: \[ \lim_{x \to 9} (\sqrt{x} + 3) \] Now, we can simply substitute \(x = 9\) into the simplified expression: \[ \lim_{x \to 9} (\sqrt{x} + 3) = \sqrt{9} + 3 = 3 + 3 = 6 \] Therefore, the solution to the limit is \(\boxed{6}\). |
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| deepseekr1 | To evaluate the limit \(\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}\), we first recognize the indeterminate form \(\frac{0}{0}\) when substituting \(x = 9\). To resolve this, multiply the numerator and denominator by the conjugate of the denominator, \(\sqrt{x} + 3\): \[ \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} = \frac{(x - 9)(\sqrt{x} + 3)}{x - 9}. \] Canceling the common factor \((x - 9)\) (valid since \(x \neq 9\)), the expression simplifies to: \[ \sqrt{x} + 3. \] Taking the limit as \(x \to 9\): \[ \lim_{x \to 9} (\sqrt{x} + 3) = \sqrt{9} + 3 = 3 + 3 = 6. \] **Answer:** \(\boxed{6}\) |
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| gpto1 | To evaluate the limit: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \] we notice that direct substitution yields a **\( \frac{0}{0} \)** indeterminate form: - As \( x \to 9 \), the numerator \( x - 9 \to 0 \). - The denominator \( \sqrt{x} - 3 \to \sqrt{9} - 3 = 3 - 3 = 0 \). This indicates that we need to simplify the expression or apply L'Hôpital's Rule. Below are two methods to find the limit. --- **Method 1: Algebraic Simplification** We can simplify the expression by factoring the numerator. **Step 1: Factor the Numerator** Notice that \( x - 9 \) can be expressed using the difference of squares: \[ x - 9 = (\sqrt{x})^2 - 3^2 = (\sqrt{x} - 3)(\sqrt{x} + 3) \] **Step 2: Simplify the Fraction** Substitute the factored form back into the expression: \[ \frac{x - 9}{\sqrt{x} - 3} = \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{\sqrt{x} - 3} \] Since \( \sqrt{x} \ne 3 \) in the neighborhood around \( x = 9 \) (except at \( x = 9 \)), we can cancel \( \sqrt{x} - 3 \): \[ \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{\sqrt{x} - 3} = \sqrt{x} + 3 \] **Step 3: Evaluate the Limit** Now, as \( x \to 9 \): \[ \lim_{x \to 9} \sqrt{x} + 3 = \sqrt{9} + 3 = 3 + 3 = 6 \] --- **Method 2: L'Hôpital's Rule** Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. **Step 1: Differentiate the Numerator and Denominator** - Numerator derivative: \( \frac{d}{dx}(x - 9) = 1 \). - Denominator derivative: \[ \frac{d}{dx}(\sqrt{x} - 3) = \frac{1}{2\sqrt{x}} \] **Step 2: Apply L'Hôpital's Rule** \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = \lim_{x \to 9} \frac{1}{\dfrac{1}{2\sqrt{x}}} = \lim_{x \to 9} 2\sqrt{x} \] **Step 3: Evaluate the Limit** \[ \lim_{x \to 9} 2\sqrt{x} = 2 \times \sqrt{9} = 2 \times 3 = 6 \] --- **Conclusion** Both methods yield the same result. Therefore, the limit is: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = 6 \] **Answer:** 6 |
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| GPT-4o | To find the limit \(\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}\), we start by substituting \(x = 9\) directly into the expression. \[ \frac{x - 9}{\sqrt{x} - 3} \] Substituting \(x = 9\): \[ \frac{9 - 9}{\sqrt{9} - 3} = \frac{0}{0} \] This gives us an indeterminate form \(\frac{0}{0}\). To resolve this, we need to simplify the expression. One effective method is to use algebraic manipulation, specifically rationalizing the denominator. Here's the step-by-step process: 1. Multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{x} - 3\) is \(\sqrt{x} + 3\). \[ \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} \] 2. Simplify the denominator using the difference of squares formula, \((a - b)(a + b) = a^2 - b^2\): \[ (\sqrt{x} - 3)(\sqrt{x} + 3) = (\sqrt{x})^2 - 3^2 = x - 9 \] So, the expression becomes: \[ \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} \] 3. Cancel out the common factor \(x - 9\) from the numerator and the denominator: \[ \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} = \sqrt{x} + 3 \quad \text{for} \, x \neq 9 \] 4. Now, we take the limit as \(x\) approaches 9: \[ \lim_{x \to 9} (\sqrt{x} + 3) \] 5. Substitute \(x = 9\): \[ \sqrt{9} + 3 = 3 + 3 = 6 \] Therefore, the limit is: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = 6 \] So, the solution is: \[ \boxed{6} \] |
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| llama3.1-70B-Instruct | To evaluate the limit, we can start by rationalizing the denominator: \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} We can multiply the numerator and denominator by the conjugate of the denominator, which is \sqrt{x} + 3: \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} This simplifies to: \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} Now, we can cancel out the (x - 9) terms: \lim_{x \to 9} \sqrt{x} + 3 As x approaches 9, \sqrt{x} approaches 3, so: \lim_{x \to 9} \sqrt{x} + 3 = 3 + 3 = 6 Therefore, the limit is: \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = 6 |
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| llama3.1-405B-Instruct | To find the limit of the expression as x approaches 9, we can start by rationalizing the denominator. \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} This simplifies to: \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} Now, we can cancel out the (x - 9) terms: \lim_{x \to 9} \sqrt{x} + 3 As x approaches 9, the value of \sqrt{x} approaches 3, so the limit is: \lim_{x \to 9} \sqrt{x} + 3 = 3 + 3 = 6 Therefore, the limit of the expression as x approaches 9 is 6. |
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| DeepSeek-Coder-V2 | To solve the problem \(\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}\), we need to handle the indeterminate form \(0/0\) that arises when directly substituting \(x = 9\). We can use the technique of rationalization to simplify the expression. Step-by-step solution: 1. **Identify the indeterminate form**: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \] Substituting \(x = 9\) directly gives: \[ \frac{9 - 9}{\sqrt{9} - 3} = \frac{0}{3 - 3} = \frac{0}{0} \] This is an indeterminate form, so we need to simplify the expression. 2. **Rationalize the denominator**: Multiply the numerator and the denominator by the conjugate of the denominator, \(\sqrt{x} + 3\): \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \] This gives: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} \] 3. **Simplify the denominator**: The denominator simplifies using the difference of squares: \[ (\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9 \] So the expression becomes: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} \] 4. **Cancel the common factor**: The \(x - 9\) terms cancel out: \[ \lim_{x \to 9} (\sqrt{x} + 3) \] 5. **Evaluate the limit**: Now, substitute \(x = 9\) into the simplified expression: \[ \sqrt{9} + 3 = 3 + 3 = 6 \] Therefore, the limit is: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = 6 \] The final answer is: \[ \boxed{6} \] |
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| Qwen2-72B-Instruct | To solve the limit \(\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}\), we can apply a technique known as rationalization or multiplying by the conjugate to remove the radical from the denominator. The idea is to multiply the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{x} + 3\). \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \] By doing so, we get: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} \] The denominator simplifies to a difference of squares: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} \] Since \(x \neq 9\) in the limit, we can safely cancel out the \(x - 9\) terms: \[ \lim_{x \to 9} (\sqrt{x} + 3) \] Now, we can simply substitute \(x = 9\) into the simplified expression: \[ \lim_{x \to 9} (\sqrt{x} + 3) = \sqrt{9} + 3 = 3 + 3 = 6 \] Therefore, the solution to the limit is \(\boxed{6}\). |
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| baichuan4 | To find the limit of the expression as $x$ approaches $9$, we can simplify the expression by rationalizing the numerator or denominator. In this case, we will multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x} + 3$. \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \] When we multiply through, we get: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} \] This simplifies to: \[ \lim_{x \to 9} \frac{x\sqrt{x} + 3x - 9\sqrt{x} - 27}{x - 9} \] Since $x$ is approaching $9$, we can substitute $9$ for $x$ in the expression: \[ \frac{9\sqrt{9} + 3 \cdot 9 - 9\sqrt{9} - 27}{9 - 9} \] This simplifies to: \[ \frac{27 + 27 - 27}{0} \] Which is: \[ \frac{27}{0} \] Since we cannot divide by zero, this means that the limit does not exist. Therefore, the limit of the expression as $x$ approaches $9$ is undefined. |
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| doubao | We can rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{x}+3$. \[ \begin{align*} &\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3} \times \frac{\sqrt{x}+3}{\sqrt{x}+3}\\ =&\lim_{x \to 9} \frac{(x-9)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}\\ =&\lim_{x \to 9} \frac{(x-9)(\sqrt{x}+3)}{x-9}\\ =&\lim_{x \to 9} (\sqrt{x}+3)\\ =&\sqrt{9}+3\\ =&3+3\\ =&6 \end{align*} \] |
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| minimax | To find the limit \(\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}\), we can use algebraic manipulation to simplify the expression and eliminate the indeterminate form that arises when we substitute \(x = 9\) directly into the expression. First, notice that both the numerator and the denominator approach zero as \(x\) approaches 9, which gives us an indeterminate form of \(\frac{0}{0}\). To simplify the expression, we can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{x} - 3\) is \(\sqrt{x} + 3\). So, we have: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \] Multiplying the numerator and the denominator, we get: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} \] Now, let's expand both the numerator and the denominator: Numerator: \[ (x - 9)(\sqrt{x} + 3) = x\sqrt{x} + 3x - 9\sqrt{x} - 27 \] Denominator: \[ (\sqrt{x} - 3)(\sqrt{x} + 3) = (\sqrt{x})^2 - (3)^2 = x - 9 \] Now, we can substitute \(x = 9\) into the simplified expression: \[ \lim_{x \to 9} \frac{x\sqrt{x} + 3x - 9\sqrt{x} - 27}{x - 9} \] \[ = \frac{9\sqrt{9} + 3(9) - 9\sqrt{9} - 27}{9 - 9} \] \[ = \frac{27 + 27 - 27 - 27}{0} \] \[ = \frac{0}{0} \] We still have an indeterminate form, which means we need to simplify further. Notice that the terms \(9\sqrt{9}\) and \(-9\sqrt{9}\) cancel each other out, as do \(27\) and \(-27\). This leaves us with: \[ \lim_{x \to 9} \frac{0}{x - 9} \] Since \(x - 9\) approaches 0 as \(x\) approaches 9, we can see that the numerator is 0 and the denominator is approaching 0 from the positive side (since \(x\) is approaching 9 from the right). This means the limit is 0. Therefore, the solution to the limit is: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = 0 \] |
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| kimi | To solve the limit problem: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \] We can notice that as \(x\) approaches 9, the denominator \(\sqrt{x} - 3\) approaches 0, which would make the expression undefined. However, we can try to simplify the expression to see if we can find a limit. First, let's rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \] This gives us: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} \] The denominator simplifies to: \[ (\sqrt{x} - 3)(\sqrt{x} + 3) = x - 9 \] So the expression becomes: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} \] We can cancel out the \(x - 9\) terms, assuming \(x \neq 9\), which leaves us with: \[ \lim_{x \to 9} (\sqrt{x} + 3) \] Now, we can simply substitute \(x = 9\) into the simplified expression: \[ \sqrt{9} + 3 = 3 + 3 = 6 \] So the limit is: \[ \boxed{6} \] |
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| wenxinyiyan | 首先，我们将分子和分母都乘以$(\sqrt{x} + 3)$，得到： $\lim_{{x \to 9}} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \lim_{{x \to 9}} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9}$ 由于$x - 9$在$x \neq 9$时不会等于0，我们可以约去分子和分母中的$x - 9$，得到： $= \lim_{{x \to 9}} (\sqrt{x} + 3) = \sqrt{9} + 3 = 3 + 3 = 6$ 所以，$\lim_{{x \to 9}} \frac{x - 9}{\sqrt{x} - 3} = 6$。 |
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| zhipu | To find the limit \(\lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3}\), we need to handle the indeterminate form that arises when we substitute \(x = 9\) directly into the expression, which results in \(\frac{0}{0}\). To resolve this, we can use algebraic manipulation. Let's start by multiplying the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{x} + 3\): \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} \] This simplifies to: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{(\sqrt{x} - 3)(\sqrt{x} + 3)} \] Notice that the denominator is a difference of squares, which can be simplified as follows: \[ (\sqrt{x} - 3)(\sqrt{x} + 3) = (\sqrt{x})^2 - 3^2 = x - 9 \] So the expression now becomes: \[ \lim_{x \to 9} \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} \] We can cancel out the \(x - 9\) terms in the numerator and the denominator: \[ \lim_{x \to 9} (\sqrt{x} + 3) \] Now, we can directly substitute \(x = 9\) into the simplified expression: \[ \sqrt{9} + 3 = 3 + 3 = 6 \] Therefore, the limit is: \[ \boxed{6} \] |
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| gpto1 | To evaluate the limit: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} \] we notice that direct substitution yields a **\( \frac{0}{0} \)** indeterminate form: - As \( x \to 9 \), the numerator \( x - 9 \to 0 \). - The denominator \( \sqrt{x} - 3 \to \sqrt{9} - 3 = 3 - 3 = 0 \). This indicates that we need to simplify the expression or apply L'Hôpital's Rule. Below are two methods to find the limit. --- **Method 1: Algebraic Simplification** We can simplify the expression by factoring the numerator. **Step 1: Factor the Numerator** Notice that \( x - 9 \) can be expressed using the difference of squares: \[ x - 9 = (\sqrt{x})^2 - 3^2 = (\sqrt{x} - 3)(\sqrt{x} + 3) \] **Step 2: Simplify the Fraction** Substitute the factored form back into the expression: \[ \frac{x - 9}{\sqrt{x} - 3} = \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{\sqrt{x} - 3} \] Since \( \sqrt{x} \ne 3 \) in the neighborhood around \( x = 9 \) (except at \( x = 9 \)), we can cancel \( \sqrt{x} - 3 \): \[ \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{\sqrt{x} - 3} = \sqrt{x} + 3 \] **Step 3: Evaluate the Limit** Now, as \( x \to 9 \): \[ \lim_{x \to 9} \sqrt{x} + 3 = \sqrt{9} + 3 = 3 + 3 = 6 \] --- **Method 2: L'Hôpital's Rule** Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. **Step 1: Differentiate the Numerator and Denominator** - Numerator derivative: \( \frac{d}{dx}(x - 9) = 1 \). - Denominator derivative: \[ \frac{d}{dx}(\sqrt{x} - 3) = \frac{1}{2\sqrt{x}} \] **Step 2: Apply L'Hôpital's Rule** \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = \lim_{x \to 9} \frac{1}{\dfrac{1}{2\sqrt{x}}} = \lim_{x \to 9} 2\sqrt{x} \] **Step 3: Evaluate the Limit** \[ \lim_{x \to 9} 2\sqrt{x} = 2 \times \sqrt{9} = 2 \times 3 = 6 \] --- **Conclusion** Both methods yield the same result. Therefore, the limit is: \[ \lim_{x \to 9} \frac{x - 9}{\sqrt{x} - 3} = 6 \] **Answer:** 6 |
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| deepseekv3 | Alright, I have this limit problem to solve: \[ \lim_{x \to 9} \left( \frac{x - 9}{\sqrt{x} - 3} \right) \] At first glance, if I plug in \( x = 9 \) directly, I get: \[ \frac{9 - 9}{\sqrt{9} - 3} = \frac{0}{3 - 3} = \frac{0}{0} \] Hmm, that's an indeterminate form. So, direct substitution doesn't work here. I need to find another way to evaluate this limit. **Step 1: Simplify the Expression** I recall that when dealing with limits that result in \( \frac{0}{0} \), one common technique is to simplify the expression algebraically. Here, both the numerator and the denominator are expressions involving \( \sqrt{x} \), so maybe I can rationalize the denominator or factor something. Looking at the numerator \( x - 9 \), I notice that it can be written as a difference of squares: \[ x - 9 = (\sqrt{x})^2 - 3^2 = (\sqrt{x} - 3)(\sqrt{x} + 3) \] Wait, that's interesting! So, the numerator can be factored into \( (\sqrt{x} - 3)(\sqrt{x} + 3) \). **Step 2: Factor the Numerator** Let's rewrite the original expression using this factorization: \[ \frac{x - 9}{\sqrt{x} - 3} = \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{\sqrt{x} - 3} \] **Step 3: Cancel Common Terms** Now, I see that \( \sqrt{x} - 3 \) appears in both the numerator and the denominator. As long as \( \sqrt{x} - 3 \neq 0 \), which it isn't when \( x \to 9 \) (since \( \sqrt{9} = 3 \)), I can cancel these terms: \[ \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{\sqrt{x} - 3} = \sqrt{x} + 3 \] **Step 4: Evaluate the Simplified Limit** Now, the expression simplifies to \( \sqrt{x} + 3 \). So, the limit becomes: \[ \lim_{x \to 9} (\sqrt{x} + 3) \] This is much simpler! I can now substitute \( x = 9 \) directly: \[ \sqrt{9} + 3 = 3 + 3 = 6 \] **Step 5: Conclusion** Therefore, the value of the limit is: \[ \boxed{6} \] **Reflection** At first, I was stuck because direct substitution led to an indeterminate form. However, by recognizing that the numerator could be factored as a difference of squares involving \( \sqrt{x} \), I was able to simplify the expression and cancel out the problematic term. This approach made the limit straightforward to evaluate. It's a good reminder that algebraic manipulation is often key to solving limit problems that initially seem tricky. |
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