Calculus Clarification on L'Hopital's Rule?

Question

I wrote:

L’Hopital’s rule is generally used to evaluate indeterminate forms 0/0. (Thomas, Weir, Hass, Giordano, 2007)
When the f(x) and g(x) limits exist for impending something in a certain manner, the limit for f(x) + g(x) exist and is the sum of the limits, this goes for subtraction and multiplication. If the limits for both of the two functions are either zero or infinite then the approach has no clear-cut answer, and for that reason would not be correct. (Thomas, Weir, Hass, Giordano, 2007)

[f(x)/g(x)] = lim [f'(x)/g'(x)]

In this case we could apply L’Hopital’s rule, in most scenarios the two derivatives will have a distinct non-zero limit which makes it easier in the end to apply L’Hopital’s rule. If f’(x) and g’(x) have limits which are both either zero or infinite the rule will still actually work. This would result in having to apply the rule multiple times to come to an end result. (Thomas, Weir, Hass, Giordano, 2007)
However, if the f(x) and g(x) do not meet the condition just mentioned, the rule would give an incorrect result if,
F(x) = 1 and g(x) = x

The limit of f/g as x gets closer to 1 would be 1/1 = 1
L’Hopital’s rule would also give incorrect results if f’(x) = 0 and g’(x) = 1. This is because the limit for f'/g' = 0/1 = 0. (Thomas, Weir, Hass, Giordano, 2007)

They sent it back to me saying:
Comments on this criterion: The work states that "if the limits for both of the two functions are either zero or infinite," then "... in this case we could apply l'Hopital's rule." However this is insufficient. The following points require clarification:

1. It is not clear what expression "both" functions should be in, since the same paragraph in which this statement occurs has the expression "f(x) + g(x)" while elsewhere in the essay there is the expression f(x)/g(x) and sometimes f'(x)/g'(x).

2. It is not clear what l'Hopital's rule itself is. What is the "rule" that one can apply if it turns out that "both... functions are zero or infinite?"

3. The author has not identified all of the preconditions which must be met in order to apply the rule. Though "both... functions (must be) either zero or infinit(y)," this is not the only condition which must be met.

The thing is I copied it word for word from the book so I dont really understand what is going on. Can someone point me in the right direction on how to revise this? I am crunched for time, my term ends in 2 months and I am so behind from having two calculus classes in one term!

Answer 1

There are several problems with this that I notice. First, other than the first sentence, the first paragraph has nothing to do with L'Hopital's Rule. The fact that lim (x-->a) [f(x) + g(x)] exists when lim (x-->a) f(x) and lim (x-->a) g(x) both exist and are finite is irrelevant to L'Hopital's Rule.



Secondly, you have stated that the rule is generally used to calculate 0/0 limits, but this is only half of what L'Hopital's Rule can do; L'Hopital's Rule can also compute limits of the form ±infinity/infinity (that is: f(x) --> -infinity or infinity as g(x) --> -infinity or infinity). You've mentioned a little bit about this in your paper, but you didn't make this point clear.



Third, like one of your comments, you didn't really state how the rule is used. Basically, if we have a limit of the form lim (x-->a) f(x)/g(x), where f(x)/g(x) takes the form 0/0 or ±infinity/infinity, and f'(x) and g'(x) exist near x = a, then:

lim (x-->a) f(x)/g(x) = lim (x-->a) f'(x)/g'(x).



Combine this is what you have bout this and you should be fine in this regard.



I hope this helps!

Answer 2

L'Hopital's rule tells us the following: If lim f(x) = 0 and lim g(x) = 0, lim f(x)/g(x) = lim f'(x)/g'(x) (this is as all as x goes to some number c)



For example, consider lim x^2/x as x goes to 0. This can be written as lim f(x)/g(x) with f(x)=x^2 and g(x)=x.



lim x^2 as x goes to 0 is 0 since 0^2 = 0. lim x as x goes to 0 if of course 0, therefore we can apply L'Hopital's rule, that is, lim f(x)/g(x) = lim f'(x)/g'(x) = lim 2x/1 (since the f'(x)=d/dx[x^2] = 2x and g'(x) = d/dx[x] = x).



lim 2x/1 as x goes to 0 is of course 1 by substituting 0 into 2x/1 and so by L'Hopital's rule, lim x^2/x = 0.



L'Hopital's rule also says that if lim f(x) = infinity and lim g(x) = infinity, again, lim f(x)/g(x) = f'(x)/g'(x).



If you consider lim (3x^2+3)/(x^+1) as x goes to infinity, taking the limit of the top function as x goes to infinity and taking the limit of the bottom function as x goes to infinity gives us infinity/infinity.



Therefore L'Hopital's rule tells us lim (3x^2+3)/(x^+1) = lim d/dx(3x^2+3)/d/dx(x^2+1) = lim 6x/2x = 3, and so lim (3x^2+3)/(x^+1) as x goes to infinity is 3.



Hope this helps. If not, you'll need to consult a text book or Wikipedia.

Answer 3

lim (x?0) [e^x - e^-x - 2x]/[x - sinx] (type: 0/0) = lim (x?0) [e^x + e^-x - 2]/[a million - cosx] (type: 0/0) = lim (x?0) [e^x - e^-x]/[sinx] (type: 0/0) = lim (x?0) [e^x + e^-x]/[cosx] = [e^0 + e^0]/[cos0] = 2/a million = 2