Are there any situations in which L'Hopital's Rule WILL NOT work?

Question

Today was my first day learning L'Hopital's Rule, and I was wondering if there are any situations in which you cannot use this rule, with the exception of when a limit is determinable.

Answer 1

L'Hopital's Rule will work in any cases involving limits that go to 0/0 or ∞/∞. If the resultant new limit is also indeterminate, you can use L'Hopital's Rule again.

Answer 2

L'Hospital's rule can only be used when you attempt to take the limit of a quotient and obtain 0/0 which is undefined.

It is then OK to take the derivative of the numerator & denominator to obtain a meaningful answer!

Answer 3

The rule is only used when you attempt to take a limit of a quotient and the result is 0/0!

Answer 4

L' Hôspital's Rule does not help with lim (x →∞) of √(10x+1)÷√(x+1). The limit must be found some other way.

L' Hôspital's Rule also does not work with lim (x→π/2) of sec x/tan x. The result is an endless and fruitless effort. The limit must be found some other way.

Answer 5

L'Hopital's Rule works (as any Calculus student would know) only with indeterminate limits in the form of either 0/0 or ∞/∞.

Answer 6

L'Hopital's rule finds limits of quotients in which the numerator and denominator both approach 0 or both approach infinity. For something like the limit of x³ / 2^x as x approaches 2, the function is continuous at x = 2, so the limit can be found by substituting 2 for x, and you get 8/4 = 2. If you take the derivative of the numerator you get 3x² and for the denominator you get 2^x · log_e2. Plugging in 2 for x in that, you get 12/(4 log_e2), which is clearly wrong.

Consider the limit of x log_e x as x approaches 0 from above. You can write it as a quotient: log_e x/(1/x), and then the numerator approaches –∞ and the denominator approaches +∞, so you can apply L'Hopital's rule. But if you instead write it as x/(1/log_e x), when you differentiate the numerator, you get 1 and with the denominator you get –1/(log_e x)²·(1/x) and that simplifies to –x/(log_e x)² and you still have infinity over infinity.

With x⁵ e^–x as x to +∞, you can write it as x⁵/e^x and then it takes several iterations of L'Hopital's rule to get it to 120/e^x, and that limit is clearly 0. But although L'Hopital's rule is often very efficient and can often treat otherwise seemingly intractable limits, it gives little or no insight. There is a common-sense reason why this limit must be 0, but L'Hopital's rule does not tell you that.

One place where the use of L'Hopital's rule is quite illogical is when you're evaluating a derivative by using a limit of a difference quotient. You have ƒ'(a) = lim_{x -> a} (ƒ(a+h) – ƒ(a))/h. Here you have the numerator and the denominator both approaching 0. L'Hopital's rule will usually give you a correct result, but that misses the point. These limits are used for proving that the usual rules for differentiation are correct. But you would be using those very rules in applying L'Hopital's rule, just as if they had _already_ been proved to be correct.