Q.50 The value of lim (2x+6)/(x+3) is _____ .
x–>-3
Common Misconception
The limit \( \lim_{x \to -3} \frac{2x+6}{x+3} \) does not fail to exist as a finite number just because the expression is undefined at \( x = -3 \); the denominator becomes zero while the numerator is non-zero there, but limits concern behavior near the point, not at it.
This tests understanding of removable discontinuities in rational functions, common in calculus exams like CSIR NET.
Step-by-Step Solution
1. Simplify Numerator
The numerator factors as \( 2x + 6 = 2(x + 3) \).
2. Rewrite Fraction
Substitute: \( \frac{2x+6}{x+3} = \frac{2(x+3)}{x+3} \).
3. Cancel Common Factor
For \( x \neq -3 \), cancel \( (x+3) \) to get \( \frac{2(x+3)}{x+3} = 2 \).
4. Evaluate Limit
The simplified function equals 2 everywhere except at \( x = -3 \), so \( \lim_{x \to -3} \frac{2x+6}{x+3} = 2 \).
Multiple-Choice Options Analysis
| Option | Value | Explanation |
|---|---|---|
| A | 0 | Incorrect: Numerator approaches 0 but factors cancel. |
| B | 2 | Correct: Simplified expression is constantly 2 near \( x = -3 \). |
| C | Does not exist | Misleading: Limit exists despite undefined function value at point. |
| D | \( \infty \) or \( -\infty \) | Incorrect: Removable discontinuity, not vertical asymptote. |
Graphical Understanding
The graph of \( y = \frac{2x+6}{x+3} \) matches \( y = 2 \) everywhere except a hole at \( (-3, 2) \), confirming the limit is 2.
Key Exam Tips
- Always try direct substitution first to identify indeterminate forms like 0/0.
- For rational functions, factor numerator and denominator before advanced methods.
- “Does not exist” traps confuse function value with limit behavior.
- Removable discontinuities allow finite limits despite undefined points.
