Do you always need to go to l’Hôpital?

Are you limited when you are evaluating limits? Do you always need to go to l’Hôpital?

NoLopital -- 0.png
{Image: derived from Wiki Commons}

First of all, Seasons Greetings to all readers and my followers! I am always happy to have new followers on Steemit, even when I am not so active. Welcome everyone!

I got this calculus problem (see below) from a Facebook acquaintance of mine, who solved it using l’Hôpital’s Rule. Abhijeet is a young and talented student who loves mathematics and science, and he enjoys sharing his knowledge to others. He could be our future Einstein!

NoLopital -- q.png

I shall now demonstate my short and sweet solution, which takes only 5 steps. In mathematics, we should not just get the answer, but look for alternative approaches, and even strife for elegance and simplicity. Note that it may not be that easy to think of a 5 step solution that can be written (literally) on the back of an envelop in less than 5 minutes. That is because that may take 50 years of perfection to achieve that standard. For some people, at least. Maybe for future Einsteins, that may take less than 5 hours.

Here goes ...

NoLopital -- s.png

In the first step, we let $u=x-\frac{\pi}{2}$ , so that $x=u-\frac{\pi}{2}$ , which means $2x=2u+\pi$ .

This leads to the 2nd step. Here we use $\tan(\theta+\pi)=\tan(\theta)$ to simplify to step 3.

Now for the 3rd step ... This is the crucial bit: The $tan{2u}$ (highlighted in yellow) can be split as $\frac{\sin{2u}}{\cos{2u}}$ . To match the $\sin{2u}$ , we introduce a ‘2’ (shown in red) to join with the u in the denominator, forming the fraction $\frac{\sin{2u}}{2u}$ to prepare for the next step. But in order that the expression remains equal to the previous step’s expression, we need a counter-balancing ‘2’ in the numerator (shown also in red). Thus we have $\frac{2}{\cos{2u}}$ in the other limit.

In the fourth step, $\lim_{u\to0}\frac{\sin{2u}}{2u}=1$ using the famous limit $\lim_{\theta\to0}\frac{\sin{\theta}}{\theta}=1$ . So now you see the purpose for introducing the red ‘2’ in the previous step. The other limit (the one on the right) gives no problem, because $\frac{2}{\cos{2u}}$ is continuous at 0, which means we can just substitute u = 0, and this gives us $\frac{2}{1}$ , or 2.

It is all easy from here onwards, and we get 2 as the final answer.

In French “l’Hôpital” can mean a “hospital” or “hostel”, or both! For example, l’Hôpital des Invalides is an area in Paris for retired soldiers. Words like “hospital”, “hostel”, “hotel”, “hospice” all come from the same root.

To answer the question that is the title of this article, obviously we do not always need to use l’Hôpital’s Rule, although it is not ... er ... invalid to do so. The Rule is named after the mathematician Guillaume de l'Hôpital and it can be used, for example, when both the numerator and denominator tend to zero in a limit calculation. I avoided using this rule by using $\lim_{\theta\to0}\frac{\sin{\theta}}{\theta}=1$ . To apply this formula, we sometimes need to make some factor adjustments.

I hope you have learned something awesome in this article.

You might also want to read some of my articles below:
"Singapore Math" : Square in Quadrant
Arithmetic Progression in a triangle

If you find my articles useful or interesting, please upvote and resteem them! Thanks !

Cheers!!!

@tradersharpe
-- promoting sharp minds