Question
(12) \( \lim _{x \rightarrow 3} \frac{x^{2}-1}{x+1} \)
Ask by Li John. in Liberia
Mar 10,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The limit is 2.
Solution
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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The Deep Dive
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: \[ \lim_{x \rightarrow 3} \frac{x^{2}-1}{x+1} = 2 \]