- Topic
- Limits
- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Limits practice question 6
Which statement about the limit of the function below as is correct?
The bars around follow a different formula on each side of 3. Work out each one-sided limit on its own, then compare.
Factor the numerator as . Then use for , and for .
Theory UBC Math 100 §2.1CLP-1 §1.3Key results
Worked solution
Strategy
An absolute value has a different formula on each side of the point where its inside is zero. Rewrite the function on each side of 3 without the bars, cancel, and compare the one-sided limits.
Working
Factor the numerator; it shares the factor x − 3 with the bars.
On the right, |x − 3| = x − 3.
On the left, |x − 3| = −(x − 3), which brings a minus sign.
The one-sided limits are −5 and 5. They disagree, so the two-sided limit does not exist.
Answer
The answer is the left-hand limit is −5, the right-hand limit is 5, and the two-sided limit does not exist.
Check
f(2.99) = −4.99 and f(3.01) = 5.01, one value on each side of 3, so the two sides disagree.
Where this goes wrong
Dropping the bars and writing |x − 3| = x − 3 everywhere, which gives 5 from both sides. Left of 3 the inside is negative, so |x − 3| = 3 − x and the left-hand limit is −5.
Leans on
Absolute value as a piecewise function, from Functions, graphs and inverses
Try a similar question
Evaluate the one-sided limit.Saved in this browser only.