I think you will find this to be quite impossible without 4x l'Hopital's Rule$\displaystyle \lim_{x\to 0} \frac{(x\sin x)^2}{(x\cos x)^2-\sin^2x}$D'you have examples where we HAVE to use it ? And can't use anything else ?
The answer is $\displaystyle -\frac{3}{2}$ if you're interested.
Bad example. Because $\displaystyle \sin x$ 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 $\displaystyle \lim_{x\to 0}\frac{\sin x}{x}$ 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?
I do quite impossible for breakfast:
$\displaystyle \lim_{x\to 0} \frac{(x\sin x)^2}{(x\cos x)^2-\sin^2x}$
$\displaystyle = \lim_{x\to 0} \frac{x^2(x - x^3/6 + x^5/120 - ....)^2}{x^2(1 - x^2/2 + x^4/24 - ....)^2 -(x - x^3/6 + x^5/120 - ....)^2}$
$\displaystyle = \lim_{x\to 0} \frac{x^2(x - x^3/6 + x^5/120 - ....)^2}{x^2(1 - x^2/2 + x^4/24 - ....)^2 - x^2(1 - x^2/6 + x^4/120 - ....)^2}$
$\displaystyle = \lim_{x\to 0} \frac{(x - x^3/6 + x^5/120 - ....)^2}{(1 - x^2/2 + x^4/24 - ....)^2 - (1 - x^2/6 + x^4/120 - ....)^2}$
$\displaystyle = \lim_{x\to 0} \frac{x^2(1 - x^2/6 + x^4/120 - ....)^2}{(1 - x^2 + x^4/3 - ....) - (1 - x^2/3 + 2x^4/45 - ....)}$
$\displaystyle = \lim_{x\to 0} \frac{x^2(1 - x^2/6 + x^4/120 - ....)^2}{- x^2 + x^4/3 + x^2/3 - 2x^4/45 + ....}$
$\displaystyle = \lim_{x\to 0} \frac{x^2(1 - x^2/6 + x^4/120 - ....)^2}{x^2(- 1 + x^2/3 + 1/3 - 2x^2/45 + ....)}$
$\displaystyle = \lim_{x\to 0} \frac{(1 - x^2/6 + x^4/120 - ....)^2}{- 1 + x^2/3 + 1/3 - 2x^2/45 + ....}$
$\displaystyle = \frac{1}{-1 + 1/3} = - \frac{3}{2}$.
ps: It's totally impossible that I sometimes struggle with.
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.
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