Question
(a) \( \lim _{x \rightarrow 3^{+}} \frac{x-1}{x-3} \)
Ask by Weston Hills. in the United States
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The limit is positive infinity.
Solution
To evaluate the limit:
\[
\lim_{x \rightarrow 3^{+}} \frac{x - 1}{x - 3}
\]
**Step 1: Analyze the Behavior as \( x \) Approaches 3 from the Right**
- **Numerator (\( x - 1 \))**:
- As \( x \rightarrow 3^{+} \), \( x - 1 \) approaches \( 2 \).
- **Denominator (\( x - 3 \))**:
- As \( x \rightarrow 3^{+} \), \( x - 3 \) approaches \( 0 \) from the positive side (since \( x \) is slightly greater than 3).
**Step 2: Determine the Sign of the Fraction**
- The numerator approaches a positive value (\( 2 \)).
- The denominator approaches a small positive number (\( 0^{+} \)).
Therefore, the entire fraction:
\[
\frac{x - 1}{x - 3} \rightarrow \frac{2}{0^{+}} = +\infty
\]
**Conclusion:**
\[
\lim_{x \rightarrow 3^{+}} \frac{x - 1}{x - 3} = +\infty
\]
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Simplify this solution The Deep Dive
To evaluate the limit \( \lim _{x \rightarrow 3^{+}} \frac{x-1}{x-3} \), we first consider what happens as \( x \) approaches 3 from the right (values slightly greater than 3). The expression can be rewritten as: \[ \frac{x-1}{x-3} \] As \( x \) approaches 3, the numerator \( x - 1 \) approaches 2 (since \( 3 - 1 = 2 \)), and the denominator \( x - 3 \) approaches 0 from the positive side (since we are approaching from the right, \( x - 3 \) will be a small positive number). Thus, we have: \[ \lim _{x \rightarrow 3^{+}} \frac{x-1}{x-3} = \frac{2}{0^{+}} = +\infty \] So, the limit is \( +\infty \).
