Rules to apply L'Hospital on a limit

[Krushnaraj Pandya]

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.

 

[Krushnaraj Pandya]

(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?

 

[Krushnaraj Pandya]

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 :)

 

[Krushnaraj Pandya]

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]

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 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 $x^2.$

 

[Krushnaraj Pandya]

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.

 

[Krushnaraj Pandya]

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

 

[Krushnaraj Pandya]

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]

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]

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

I can't think of a general one off hand, but for this last one, consider differentiating the left hand side which equals ##\frac {1}{1+x}=1-x+x^2/2-....## since the lh side is a gp with ratio x , then integrate term-by-term. A bit contrived maybe, but I can remember it.