- Topic
- L’Hôpital’s rule
- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 3
L’Hôpital’s rule practice question 18
The functions and have continuous derivatives everywhere, with , , , , and . Evaluate the limit.
The bottom is , not . As , at which input is evaluated, and what form does the limit have?
Substitute first: the top tends to and the bottom to , so the form is and the rule applies.
The derivative of is by the chain rule. At , which value of does that need?
A fraction, or a decimal to three decimal places.
Theory UBC Math 100 §6CLP-1 §3.7Key results
Worked solution
Strategy
Only values are given, so the rule does all the work: check the form with f(2) and g(6), differentiate with the chain rule on g(3x), and substitute values from the list.
Working
f and g are continuous; at x = 2 the bottom is evaluated at 3x = 6. Form 0/0.
Chain rule on the bottom: the inner function 3x contributes the factor 3.
The derivatives are continuous, so substitute; g′ is needed at 6, not at 2.
Answer
The answer is four fifteenths.
Check
Near x = 2, f(x) ≈ 4(x − 2) and g(3x) ≈ 5(3x − 6) = 15(x − 2), so the quotient is close to 4/15.
Where this goes wrong
Using the values at 2 for the bottom: g(3x) needs g and g′ at 3x = 6, so g(2) and g′(2) are not used at all. Forgetting the chain-rule factor 3 is the other slip, which gives 4/5.
Leans on
The chain rule, from The chain rule and logarithms
Try a similar question
Evaluate the limit.Saved in this browser only.