Limits requiring L'Hopital's Rule

In this video showed how and when to use L'Hopital's rule ffor taking limits

[icafe]

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[tomtomspa]

the reason you can bring the limit at exponent is not because e is constant, but rather that the exp function is continuos.

[calebo]

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[ianrobinson]

Very clear demo. And unexpected answer!

I would have eased the approach by setting the expression to y and taking logs before applying l’Hôpitals rule. Then exponentiate once -1/2 is derived. It’s also not necessary in practice to keep writing the limit or derivative expression on top and bottom each time. Just go direct. But I accept you’re seeking clarity for demo purposes.

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[codeescape]

00:08 Limits requiring L'Hopital's Rule

01:55 Applying L'Hopital's Rule to find the limit of the given expression.

03:40 To evaluate the limit, we use L'Hopital's Rule.

05:56 L'Hopital's Rule can be applied to rational expressions if the limit gives zero over zero.

07:53 L'Hopital's Rule allows you to differentiate the top and bottom separately when you have a rational function with a limit of zero over zero.

09:47 L'Hopital's Rule allows us to find the limit by differentiating the numerator and denominator separately.

11:54 The application of L'Hopital's Rule is discussed in finding the limit as x approaches zero.

14:23 The limit of negative secant squared x as x approaches zero is e to the negative 1/2.

[chaddest]

There is a much simpler method for such questions. Basically, if the limit is of 1^infinity form, you can write as e^[(base - 1) * power]. After doing that, we can write cosx - 1 as x^2/2 as per the expansion series of cosx. And we instantly get the answer. However, i do acknowledge your solution as well because it's just not about finding the solution, it's also about exploring various methods. Thank you.

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[keithrobinson]

Those are very large ees! Love the video. Your eeeexplanations areee aweeesomeee!

[klementhajrullaj]

You can do it even more briefly, without using the second derivative, because the limit when x goes to 0 of tanx/x is the same as the limit when x goes to 0 of sinx/x, so equal to 1, therefore, all the limit will be: e^(-1/2)=1/e^(1/2)=1/Ve=Ve/e.

[narjsseettegrini]

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[ahmedabdelkoui]

From because lim(sinx/x )=1. We clearly understand lim in this limiting case when x tends towards zero. Consequently, without going to the tangent and the secant, the limit sought is e^(-1/2)=1/√e.

[바르고고운말]

L'hopital is very powerful. but you don't need to use that here. there is another solution. in korea, we learn this problem like below.

let cosx-1=t, then cosx=t+1, and when x->0, then t->0
(lncosx)/(x^2)


(because lim[x->0](cosx-1)/(x^2)=-1/2, lim[t->0]ln(1+t)/t=1)

i am sorry i am not good at english😢

[mark]

We can also use the fact that sinx/x approaches 1 instead of using the L'Hopital's rule for a second time

[mr.mxyzptlks]

Not so simple a question. L’Hospital’s rule is from the 16 hundreds. Complex numbers came about in the 17 hundreds. Can L’Hospital apply to C? I never learned about any of this beyond R^1, but I think it probably applies to R^n, comments invited. I don’t think it can be applied to C though. More comments, or references, or proofs, invited 🤓

[BinaHejazi]

didn't we say that ONE to ANY (repeating ANY) power is ONE?

[ЕвгенияЛысенко-ун]

Isn’t it also for infinity over infinity?