# Rules to apply L'Hospital on a limit

Gold Member

## Homework Statement

## \lim x-0 \frac {xcosx-log(1+x)}{x^2}##

## Homework Equations

##\frac{log(1+x)}{x}=1## ...(i)

## The Attempt at a Solution

Using (i) we can write numerator as xcosx-x, cancelling x from denominator we have cosx-1/x, this is 0/0 form so we can use LHR which gives us -sinx/1 but this gives an incorrect value for the limit. Is there some technicality I'm overlooking about applying LHR?

member 587159

## Homework Statement

## \lim x-0 \frac {xcosx-log(1+x)}{x^2}##

## Homework Equations

##\frac{log(1+x)}{x}=1## ...(i)

## The Attempt at a Solution

Using (i) we can write numerator as xcosx-x, cancelling x from denominator we have cosx-1/x, this is 0/0 form so we can use LHR which gives us -sinx/1 but this gives an incorrect value for the limit. Is there some technicality I'm overlooking about applying LHR?

(i) is false. There should be a limit sign. And even then, you can't substitute a limit in another limit. Do as you are supposed to do:

take derivative of numerator and denominator and then take the limit. If necessary, repeat.

Gold Member
(i) is false. There should be a limit sign. And even then, you can't substitute a limit in another limit. Do as you are supposed to do:

take derivative of numerator and denominator and then take the limit. If necessary, repeat.
I have seen sinx/x taken to be one inside a larger limit, when can we use this and when not?

Gold Member
take derivative of numerator and denominator and then take the limit. If necessary, repeat.
That is easy enough, I'm in search of quicker methods which are practical in a time bound exam

member 587159
I have seen sinx/x taken to be one inside a larger limit, when can we use this and when not?

It would be applicable if you didn't make another mistake. You split up the limits like this:

$$\lim_{x \to 0} \frac{x\cos x - \log(1+x)}{x^2} = \lim_{x \to 0} \frac{\cos x}{x} - \lim_{x \to 0} \frac{\log(1+x)}{x^2}$$

but this is only allowed when the two limits on the right exist, and they don't. You can't apply l'Hopital's rule on the first limit on the right, as filling ##x=0## in yields ##1/0## (and it should be ##0/0## to apply that rule).

Instead, apply l'Hopital's rule on the entire thing. Sometimes, wanting to go faster, makes you a lot slower :)

Gold Member
It would be applicable if you didn't make another mistake. You split up the limits like this:

$$\lim_{x \to 0} \frac{x\cos x - \log(1+x)}{x^2} = \lim_{x \to 0} \frac{\cos x}{x} - \lim_{x \to 0} \frac{\log(1+x)}{x^2}$$

but this is only allowed when the two limits on the right exist, and they don't. You can't apply l'Hopital's rule on the first limit on the right, as filling ##x=0## in yields ##1/0## (and it should be ##0/0## to apply that rule).

Instead, apply l'Hopital's rule on the entire thing. Sometimes, wanting to go faster, makes you a lot slower :)
True :D I suppose that's the way to go after all. Thank you very much for your help.

Mark44
Mentor

## Homework Statement

## \lim x-0 \frac {xcosx-log(1+x)}{x^2}##
Instead of writing \lim x-0, as you did above and in another thread, use \lim_{x \to 0}. When rendered this becomes ##\lim_{x \to 0}##.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

## \lim x-0 \frac {xcosx-log(1+x)}{x^2}##

## Homework Equations

##\frac{log(1+x)}{x}=1## ...(i)

## The Attempt at a Solution

Using (i) we can write numerator as xcosx-x, cancelling x from denominator we have cosx-1/x, this is 0/0 form so we can use LHR which gives us -sinx/1 but this gives an incorrect value for the limit. Is there some technicality I'm overlooking about applying LHR?

If l'Hospital gives ##\frac 0 0## you need to apply it again.

Alternatively, you can expand the numerator in powers of ##x##, but you need to go to terms or at least ##x^2##; when you took ##\ln(1+x) \approx x## that was not good enough. Instead, use
$$\ln(1+x) = x - \frac 1 2 x^2 + \cdots,$$
where ##\cdots## stands for terms in ##x^3, x^4, \ldots## You can stop at ##\cos x = 1 + \cdots## because when you multiply by ##x## you will be looking at the main terms of order ##x##, and the missing terms will be of order higher than ##x^2.##

Last edited:
bhobba
Gold Member
Instead of writing \lim x-0, as you did above and in another thread, use \lim_{x \to 0}. When rendered this becomes ##\lim_{x \to 0}##.
Oh right, thank you for informing me of the correct way to write it.

Gold Member
If l'Hospital gives ##\frac 0 0## you need to apply it again.

Alternatively, you can expand the numerator in powers of ##x##, but you need to go to terms or at least ##x^2##; when you took ##\ln(1+x) \approx x## that was not good enough. Instead, use
$$\ln(1+x) = x - \frac 1 2 x^2 + \cdots,$$
where ##\cdots## stands for terms in ##x^3, x^4, \ldots## You can stop at ##\cos x = 1 + \cdots## because when you multiply by ##x## you will be looking at the main terms of order ##x##, and the missing terms will be of order higher than ##x^2.##
Expansions can be used in almost any limit problem I see but there are about ten standard functions and all their expansions are so similar that it's easy to mix them up and end up with a negative in an exam

Gold Member
If l'Hospital gives ##\frac 0 0## you need to apply it again.

Alternatively, you can expand the numerator in powers of ##x##, but you need to go to terms or at least ##x^2##; when you took ##\ln(1+x) \approx x## that was not good enough. Instead, use
$$\ln(1+x) = x - \frac 1 2 x^2 + \cdots,$$
where ##\cdots## stands for terms in ##x^3, x^4, \ldots## You can stop at ##\cos x = 1 + \cdots## because when you multiply by ##x## you will be looking at the main terms of order ##x##, and the missing terms will be of order higher than ##x^2.##
Do you have some sort of mnemonic device though? That'd be really helpful

bhobba
Mentor
Do you have some sort of mnemonic device though? That'd be really helpful

L-Hopital is one of those things some get carried away with while others remember some simple corollaries and tricks using things like the Gamma function to do it. I tend to be a bit in between and remember a few simple corollaries - but when asked to be explicit write it out in full. The easiest corollary to remember is any polynomial divided by e^(αx) will be 0. Its like that in many areas of math - some like to remember simple 'tricks', some do it from first principles. A number of math professors mentioned it to me when doing my degree - each preferred one way or the other - its purely what feels good to you.

As an example in the math challenge you had to evaluate (0 to ∞) ∫x^2*e^-αx . I wrote it out in full using L-Hopital, but using a couple of 'tricks' its dead simple. Do the change of variable x' = αx and you have 1/α^3 ∫x^2*e^-x = 1/α^3 Γ(3) = 2!/α^3 = 2/α^3. It's the kind of tricks you learn with practice, but I wouldn't put it on exam papers testing if you know this stuff.

Thanks
Bill

WWGD