APFIVE Evaluating a Limit by Factoring
All of the following statements about the limit $$\lim_{x\to3} \frac{x^2-9}{x-3}$$ are true EXCEPT:
A
Cancelling the factor $$x-3$$ in $$\frac{x^2-9}{x-3}$$ is mathematically invalid.
B
After canceling the common factor, the simplified expression becomes $$x+3$$, so the limit as $$x\to3$$ is 6.
C
Factoring $$x^2-9$$ as $$(x-3)*(x+3)$$ reveals a common factor with the denominator.
D
The original function is undefined at $$x=3$$, but the limit exists and equals 6.
Question Leaderboard
| Rank | |||||
|---|---|---|---|---|---|
| #1 | annika.h.seasholes | 2 | 2 | 0m 51s | 149 |
| #2 | cinderella_mall | 1 | 1 | 0m 00s | 100 |
| #3 | thetitaniumgamer360 | 1 | 1 | 0m 00s | 100 |
| #4 | kskim0724 | 1 | 1 | 0m 07s | 93 |
| #5 | arwa.vora2008 | 1 | 1 | 0m 11s | 89 |
| #6 | alexyu2008 | 1 | 1 | 0m 17s | 83 |
| #7 | bezgina.anfisa | 1 | 1 | 0m 17s | 83 |
| #8 | cathyphan126 | 1 | 1 | 0m 18s | 82 |
| #9 | jonathannajarro05 | 1 | 1 | 0m 19s | 81 |
| #10 | soumiksaravanan | 1 | 1 | 0m 20s | 80 |
Items per page:
10
1 – 10 of 58