- Topic
- L’Hôpital’s rule
- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
L’Hôpital’s rule practice question 21
There is exactly one value of the constant for which the limit below exists. Find that , and enter the value of the limit.
The limit exists for only one . After one application of the rule, what must the new numerator do at for the limit to have a chance?
The form is for every . After one application it becomes , whose denominator tends to 0; for a finite limit the numerator must tend to 0 too.
Set equal to 0 at to find . Then apply the rule once more.
The value of the limit, as a whole number.
Theory UBC Math 100 §6CLP-1 §3.7Key results
Worked solution
Strategy
The first application produces a quotient with a vanishing denominator. Unless its numerator also vanishes, that quotient blows up with opposite signs on the two sides of 0, and so does the original, so there is no limit. That condition pins down c; a second application then gives the value.
Working
First application.
Otherwise the form is (non-zero)/0: the new quotient tends to +∞ on one side of 0 and −∞ on the other. The rule holds for one-sided and infinite limits, so the original quotient does the same, and its limit does not exist.
Second application with c = −2.
Answer
The answer is c is negative 2 and the limit is 2.
Check
With c = −2 the quotient is about 2.0134 at x = 0.01, near 2, while with c = −1 it is already about 102 there.
Where this goes wrong
Applying the rule a second time without first fixing c, which “gives” 4e2x/2 → 2 for every c. The second application is only legal when the intermediate form is 0/0, which is exactly the condition that determines c.
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