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Prove \( \lim _{x \rightarrow 5} f\left(g(x)=f\left(\lim _{x \rightarrow 5} g(x)\right)\right. \) if \( f(x)=x^{2}-3 x+2 \) and \( g(x)=2 x-3 \)

Ask by Maxwell Logan. in the United States
Feb 26,2025

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Answer

\( \lim_{x \rightarrow 5} f(g(x)) = f\left(\lim_{x \rightarrow 5} g(x)\right) = 30 \).

Solution

Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow 5}\left(gx\right)\) - step1: Rewrite the expression: \(g\times \lim _{x\rightarrow 5}\left(x\right)\) - step2: Calculate: \(g\times 5\) - step3: Calculate: \(5g\) Calculate the limit \( \lim_{x \rightarrow 5} f(g(x)) \). Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow 5}\left(fgx\right)\) - step1: Rewrite the expression: \(fg\times \lim _{x\rightarrow 5}\left(x\right)\) - step2: Calculate: \(fg\times 5\) - step3: Calculate: \(5fg\) Substitute \( x=5 \) into the expression \( 2x-3 \). Evaluate the expression by following steps: - step0: Evaluate: \(2x-3\) - step1: Substitute: \(2\times 5-3\) - step2: Multiply the numbers: \(10-3\) - step3: Subtract the numbers: \(7\) Substitute \( x=5 \) into the expression \( x^{2}-3x+2 \). Evaluate the expression by following steps: - step0: Evaluate: \(x^{2}-3x+2\) - step1: Substitute: \(5^{2}-3\times 5+2\) - step2: Multiply the numbers: \(5^{2}-15+2\) - step3: Evaluate the power: \(25-15+2\) - step4: Calculate: \(12\) To prove the limit \( \lim _{x \rightarrow 5} f\left(g(x)\right) = f\left(\lim _{x \rightarrow 5} g(x)\right) \), we will follow these steps: 1. **Calculate \( \lim_{x \rightarrow 5} g(x) \)**: - Given \( g(x) = 2x - 3 \). - Substitute \( x = 5 \): \[ g(5) = 2(5) - 3 = 10 - 3 = 7 \] - Thus, \( \lim_{x \rightarrow 5} g(x) = 7 \). 2. **Calculate \( f(g(5)) \)**: - We need to find \( f(7) \) since \( g(5) = 7 \). - Given \( f(x) = x^2 - 3x + 2 \): \[ f(7) = 7^2 - 3(7) + 2 = 49 - 21 + 2 = 30 \] 3. **Calculate \( \lim_{x \rightarrow 5} f(g(x)) \)**: - We need to find \( f(g(x)) \) as \( x \) approaches 5. - Since \( g(x) = 2x - 3 \), we substitute \( g(x) \) into \( f \): \[ f(g(x)) = f(2x - 3) = (2x - 3)^2 - 3(2x - 3) + 2 \] - Now, we will evaluate \( \lim_{x \rightarrow 5} f(g(x)) \): - First, calculate \( g(5) = 7 \) as we did before. - Thus, we can directly substitute \( g(5) \) into \( f \): \[ \lim_{x \rightarrow 5} f(g(x)) = f(7) = 30 \] 4. **Conclusion**: - We have: \[ \lim_{x \rightarrow 5} f(g(x)) = 30 \] \[ f\left(\lim_{x \rightarrow 5} g(x)\right) = f(7) = 30 \] - Therefore, we conclude that: \[ \lim_{x \rightarrow 5} f(g(x)) = f\left(\lim_{x \rightarrow 5} g(x)\right) \] - This proves the statement.

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To prove \( \lim _{x \rightarrow 5} f(g(x)) = f\left(\lim _{x \rightarrow 5} g(x)\right) \), we first need to evaluate each component step by step. 1. **Finding \( \lim_{x \rightarrow 5} g(x) \)**: \[ g(x) = 2x - 3 \] Now, substituting \( x = 5 \): \[ g(5) = 2(5) - 3 = 10 - 3 = 7 \] Thus, \( \lim_{x \rightarrow 5} g(x) = 7 \). 2. **Finding \( f(g(x)) \)**: We know \( f(x) = x^2 - 3x + 2 \), so we want to find \( f(g(x)) \): \[ f(g(x)) = f(2x - 3) = (2x - 3)^2 - 3(2x - 3) + 2 \] Now, let’s simplify \( f(g(x)) \): - Expanding \( (2x - 3)^2 \): \[ (2x - 3)^2 = 4x^2 - 12x + 9 \] - Calculating \( -3(2x - 3) \): \[ -3(2x - 3) = -6x + 9 \] - Now combining these: \[ f(g(x)) = 4x^2 - 12x + 9 - 6x + 9 + 2 = 4x^2 - 18x + 20 \] 3. **Finding \( \lim_{x \rightarrow 5} f(g(x)) \)**: Now we evaluate: \[ \lim_{x \rightarrow 5} f(g(x)) = \lim_{x \rightarrow 5} (4x^2 - 18x + 20) \] Substituting \( x = 5 \): \[ = 4(5^2) - 18(5) + 20 = 4(25) - 90 + 20 = 100 - 90 + 20 = 30 \] 4. **Finding \( f(\lim_{x \rightarrow 5} g(x)) \)**: We now find \( f(7) \) since \( \lim_{x \rightarrow 5} g(x) = 7 \): \[ f(7) = 7^2 - 3(7) + 2 = 49 - 21 + 2 = 49 - 21 + 2 = 30 \] Finally, since both limits are equal: \[ \lim_{x \rightarrow 5} f(g(x)) = f\left(\lim_{x \rightarrow 5} g(x)\right) = 30 \] Thus the statement \( \lim_{x \rightarrow 5} f(g(x)) = f\left(\lim_{x \rightarrow 5} g(x)\right) \) is proven. The result is: \[ \boxed{30} \]

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