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@PrimeNewtons

5 ай бұрын

I appreciate that!

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

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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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@PrimeNewtons

Жыл бұрын

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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.

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From {lim(sinx/cosx)}/{lim(2x)}=lim{sinx/(2xcosx)}=lim{(sinx/x)(1/2cosx)=1/2 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.

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.

@nabilahmed7534

3 ай бұрын

Yeah

What If we use log instead ln?

Hey i have a question. If i replace the e and natural log to another constant and another logarithm, does the answer change?

@redpepper74

4 ай бұрын

It wouldn’t because when you differentiate log_n(x) you get 1/x divided by the constant ln(n). Then, because that constant is in the exponent, you can get rid of it by using the fact that n^(1/ln n) = n^(log_n(e)) = e.

Isn't limit of tanx/x is already 1, If yes, why extra step?

Optimus magister in hoc mundo !!!

Can we solve without L Hopital's rule ?

Limit of tan x/x = 1 if x tends to zero

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) =(cosx-1)(lncosx)/(cosx-1)(x^2) ={(cosx-1)/(x^2)}×{ln(t+1)/t}=-1/2 (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😢

Tanx/x = 1

That is the limit when X approaches 0 from the negative numbers, when X approaches 0 from the positive numbers, the limit is 1.

@brendanward2991

5 ай бұрын

But the function is symmetric about the y-axis. Replacing x with negative-x does not change it.

12:12 (lim)(x→0) sin⁡x/x=1