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(12) \( \lim _{x \rightarrow 3} \frac{x^{2}-1}{x+1} \)

Ask by Li John. in Liberia
Mar 10,2025

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The limit is 2.

Solución

1. Since \( x = 3 \) does not make the denominator zero, we can substitute directly into the expression. 2. Substitute \( x = 3 \) into the numerator: \[ x^2 - 1 = 3^2 - 1 = 9 - 1 = 8. \] 3. Substitute \( x = 3 \) into the denominator: \[ x + 1 = 3 + 1 = 4. \] 4. Thus, the limit becomes: \[ \lim_{x \rightarrow 3} \frac{x^2 - 1}{x+1} = \frac{8}{4} = 2. \]

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To find \( \lim _{x \rightarrow 3} \frac{x^{2}-1}{x+1} \), we can directly substitute \( x = 3 \) into the expression. Calculating the numerator: \[ 3^{2} - 1 = 9 - 1 = 8 \] Calculating the denominator: \[ 3 + 1 = 4 \] Now we substitute these values into the limit: \[ \lim _{x \rightarrow 3} \frac{x^{2}-1}{x+1} = \frac{8}{4} = 2 \] Thus, the answer is: \[ \lim_{x \rightarrow 3} \frac{x^{2}-1}{x+1} = 2 \]

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