- Topic
- Limits
- Difficulty
- Challenging
- Suggested time
- 5 min
- Marks
- 3
Limits practice question 28
There is exactly one value of the constant for which the limit below exists. Find that value of , and then enter the value of the limit.
What does the denominator do as ? If it heads to 0, what must the numerator do for the limit to have a chance of existing?
The denominator factors as , so it tends to 0. For the limit to exist the numerator must also tend to 0 at ; otherwise the quotient blows up.
Set to find , then factor the numerator, cancel , and substitute.
The value of the limit, as a fraction or a decimal.
Theory UBC Math 100 §2.1.1CLP-1 §1.4Key results
Worked solution
Strategy
When the denominator tends to zero, the limit can only exist if the numerator does too; that condition determines a. With a fixed, the limit is an ordinary factor-and-cancel.
Working
A non-zero numerator over a vanishing denominator has no finite limit.
Force the numerator to vanish at x = 3.
With a = 1 the numerator has the factor (x − 3) that cancels.
Substitute into the continuous expression that remains.
Answer
The answer is a equals 1 and the limit is seven fifths.
Check
With a = 1, at x = 3.01 the quotient is 0.0701/0.0501 ≈ 1.399, close to 7/5.
Where this goes wrong
Reporting a = 1 as the answer and stopping, or substituting x = 3 into the original quotient and getting 0/0. Finding a is half the question; the limit still has to be evaluated after the cancellation.
Leans on
Factoring a quadratic, from Functions, graphs and inverses
Try a similar question
Evaluate the limit.Saved in this browser only.