- Topic
- L’Hôpital’s rule
- Difficulty
- Challenging
- Suggested time
- 6 min
- Marks
- 4
L’Hôpital’s rule practice question 23
There is exactly one value of the constant for which the limit below is finite. Find that , and enter the value of the limit.
Why would most values of make this limit infinite? Think about what the first application of the rule leaves in the denominator.
The form is for every . After one application, cancel a factor of : .
The new denominator tends to 0, so for a finite limit the numerator must tend to 0 too. That fixes ; then apply the rule once more.
The value of the limit, as a fraction or a decimal.
Theory UBC Math 100 §6CLP-1 §3.7Key results
Worked solution
Strategy
Apply the rule once and simplify. The new denominator vanishes at 0, so the numerator must vanish too, or the limit is infinite; that condition pins down k, and one more application gives the value.
Working
Chain rule on ekx², then cancel a factor of x.
Otherwise the form is (non-zero)/0 and the limit is +∞ or −∞.
With k = 3 the form is 0/0 again; the derivative of 6e3x² is 36x e3x².
Answer
The answer is k is 3 and the limit is nine halves.
Check
With k = 3 the quotient is about 4.5005 at x = 0.01, near 9/2, while with k = 4 it is about 10,008 there.
Where this goes wrong
Applying the rule a second time before fixing k. After the first application the form is (2k − 6)/0 unless k = 3, and a second application is only allowed when the form is 0/0: that condition is what determines k.
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