# Math Help - About the real utility of l'Hôpital's rule

1. Originally Posted by angel.white
On the sin x / x , you can also use the squeeze theorem.
How about power series for $\sin x$? The answer is pretty faster, or make $\lim_{x\to0}\frac{\sin x}x=\lim_{x\to0}\int_0^1\cos(xy)\,dy,$ then as $x\to0$ the limit is 1.

2. Between mathstud who is fond of l'Hôpital's rule and you who is fond of integrals, i feel like a stranger
However, integrals are much more useful than this rule ^^'

Yep, it's very quick with power series :P

3. ## Ok

Originally Posted by Moo
It's a nonsense to use it Would you have said the same thing in English ? (just asking, i don't know if it exists ^^)

Do you use Riemann sums that often ? It's just useful because it's a comparison criteria to know if an integral or a sum diverges or converges. And sometimes, there are no other way to do it...
Sometimes, it may be nicer, for getting more rapidly to the solution... But there is no real work of thinking behind the use of this rule... Or well, i guess so.

Convince me that l'Hôpital's rule is interesting to be used and i'll trust you
In a concise and logical manner can you compute $\lim_{x \to {\infty}}(x+e^{x})^{\frac{2}{x}}$

4. oO

By taking the logarithm, and then compare the increasing (we know that ln < polynomials < exponential)

^^

5. D'you have examples where we HAVE to use it ? And can't use anything else ?
I think you will find this to be quite impossible without 4x l'Hopital's Rule $\lim_{x\to 0} \frac{(x\sin x)^2}{(x\cos x)^2-\sin^2x}$

The answer is $-\frac{3}{2}$ if you're interested.

6. Originally Posted by Mathstud28
the best example would be something like $lim_{x \to 0}\frac{sin(x)}{x}$...and knowing your style you will protest and say you need to just know its 1...but what if you didnt know that? you could do a rate of decreasing...but it is just much more in my opinion beautiful to say $\lim_{x \to 0}\frac{sin(x)}{x}=\lim_{x \to 0}\frac{cos(x)}{1}=1$...there your blatantly sardonic question is answered.
Bad example. Because $\sin x$ is formally defined in its power series.

Originally Posted by Moo
Hello,

Looking at which frequency l'Hôpital's rule is used, i'd like to know what is its real utility... D'you have examples where we HAVE to use it ? And can't use anything else ?

As a memento :

If $\lim_{x \to a} f(x)=\lim_{x \to a} g(x)=0 \text{\ or \ } \infty$, then :

$\lim_{x \to a} \frac{f(x)}{g(x)}=\lim_{x \to a} \frac{f'(x)}{g'(x)}$
I can solve basically all limits problems without using L'Hopital. There was one problem that I was working on that I really had no idea without L'Hopital.

Power series is a good method because it is almost L'Hopital but without the need of using repeated derivatives.

Originally Posted by Krizalid
How about power series for $\sin x$?
Krizalid is correct here. The limit $\lim_{x\to 0}\frac{\sin x}{x}$ is trivial because sine is defined in terms of its power series, not the informal trigonometric approach in a first semester calculus course.

@Polymerase: Taylor polynomial would be my approach .

7. ## Ok

Originally Posted by ThePerfectHacker
Bad example. Because $\sin x$ is formally defined in its power series.

I can solve basically all limits problems without using L'Hopital. There was one problem that I was working on that I really had no idea without L'Hopital.

Power series is a good method because it is almost L'Hopital but without the need of using repeated derivatives.

Krizalid is correct here. The limit $\lim_{x\to 0}\frac{\sin x}{x}$ is trivial because sine is defined in terms of its power series, not the informal trigonometric approach in a first semester calculus course.

@Polymerase: Taylor polynomial would be my approach .
Here is what I dont understand...I know how to apply the power series...I know how to apply the rate of increasing technique...but since I know how to apply those methods...then shouldn't I use the easiest method of attaining the limit? That is my whole limit...it is not comprehension or elegance I am railing against...it is just the amazing simplicity of L'hopital's rule...do you still diagree?

8. Originally Posted by Mathstud28
it is just the amazing simplicity of L'hopital's rule...do you still diagree?
The are reasons why I do not like to use it. The main reason is because I do not know the proof behind it. I never spend my time to work on understanding its proof.

9. ## Yes

Originally Posted by ThePerfectHacker
The are reasons why I do not like to use it. The main reason is because I do not know the proof behind it. I never spend my time to work on understanding its proof.
I tend to agree with not using something you dont fully understand...but I think I am close to full comprehension of the proof...so I like it =)...I am sure if you looked at it you could get it in like ten seconds though

10. Originally Posted by polymerase
I think you will find this to be quite impossible without 4x l'Hopital's Rule $\lim_{x\to 0} \frac{(x\sin x)^2}{(x\cos x)^2-\sin^2x}$

The answer is $-\frac{3}{2}$ if you're interested.
I do quite impossible for breakfast:

$\lim_{x\to 0} \frac{(x\sin x)^2}{(x\cos x)^2-\sin^2x}$

$= \lim_{x\to 0} \frac{x^2(x - x^3/6 + x^5/120 - ....)^2}{x^2(1 - x^2/2 + x^4/24 - ....)^2 -(x - x^3/6 + x^5/120 - ....)^2}$

$= \lim_{x\to 0} \frac{x^2(x - x^3/6 + x^5/120 - ....)^2}{x^2(1 - x^2/2 + x^4/24 - ....)^2 - x^2(1 - x^2/6 + x^4/120 - ....)^2}$

$= \lim_{x\to 0} \frac{(x - x^3/6 + x^5/120 - ....)^2}{(1 - x^2/2 + x^4/24 - ....)^2 - (1 - x^2/6 + x^4/120 - ....)^2}$

$= \lim_{x\to 0} \frac{x^2(1 - x^2/6 + x^4/120 - ....)^2}{(1 - x^2 + x^4/3 - ....) - (1 - x^2/3 + 2x^4/45 - ....)}$

$= \lim_{x\to 0} \frac{x^2(1 - x^2/6 + x^4/120 - ....)^2}{- x^2 + x^4/3 + x^2/3 - 2x^4/45 + ....}$

$= \lim_{x\to 0} \frac{x^2(1 - x^2/6 + x^4/120 - ....)^2}{x^2(- 1 + x^2/3 + 1/3 - 2x^2/45 + ....)}$

$= \lim_{x\to 0} \frac{(1 - x^2/6 + x^4/120 - ....)^2}{- 1 + x^2/3 + 1/3 - 2x^2/45 + ....}$

$= \frac{1}{-1 + 1/3} = - \frac{3}{2}$.

ps: It's totally impossible that I sometimes struggle with.

11. Originally Posted by angel.white
On the sin x / x , you can also use the squeeze theorem.
There's a nice geometric proof of the limit that uses the squeeze theorem too. I'll be lazy and just provide a hyperlink:

Calculus graphics -- Douglas N. Arnold

12. Originally Posted by mr fantastic
There's a nice geometric proof of the limit that uses the squeeze theorem too. I'll be lazy and just provide a hyperlink:

Calculus graphics -- Douglas N. Arnold
That cannot be called a geometric proof. Because we are assuming the area of a circle. While the area of a circle is derived from sine itself. Thus, this circle proof is actually circular. I tried to overcome it here but I do not see how to complete the argument.

13. Originally Posted by ThePerfectHacker
@Polymerase: Taylor polynomial would be my approach .
That's what is was doing this morning and it gives a satisfying result

14. Ok, i didn't see Mr F's post

This is far better than using 4 times l'Hôpital's rule...

15. Originally Posted by ThePerfectHacker
The are reasons why I do not like to use it. The main reason is because I do not know the proof behind it. I never spend my time to work on understanding its proof.
The reason I don't use it is the same reason I don't use the quotient rule.
It is unnecessary mental encumbrance. It is implicit in power series and
asymptotic methods, but they are more generally useful.

The only place that I have encountered limits that have to be done using
L'Hopital's rule is in A-level and below maths homework which tell the student to use it.

(also if you meet limits in real life you often are interested in asymptotic behaviour as well as the limit)

RonL

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