'tricky' derivatives with the limit definition.

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[manucitomx]

That was fun. However, isn’t using the Taylor expansion circular as it relies on the derivatives?
Thank you, professor.

[anshumanagrawal]

9:54 Just do what you did in the last one: Multiply and divide by the same "zero" which is in the exponent, so in this case as there is already an h in the denominator just multiply and divide by (2x+h), h(2x+h) is tending to zero and you can just do a substitution and use the limit (e^t -1)/t as t goes to 0, is equal to 1

[goodplacetostop]

11:55 To students and teachers, how is your semester going so far?

[diniaadil]

A lot of these limit tricks simply "recreate" the way we obtained derivative results such as chain rule, e.g. multiplying and dividing by a same term to use it as substitution is exactly how we think of chain rule.

[BlackTigerClaws]

RE: Exponential limit without Taylor series - If you leave the exponent in factored form (h(2x+h)), and recall the limit identity [exp(u)-1]/u -> 1 as u ->0, then you can build an expression to match the exponent in the numerator by multiplying by (2x+h) in the numerator and denominator.

[airtonrampim]

9:51 You can multiply the numerator and the denominator by exp(-xh - h²/2)/(x + h/2). With that, you'll get 2(x + h/2)exp(xh + h²/2)sinh(xh + h²/2)/(xh + h²/2). Now we have the limit lim(x->0) sinh(x)/x instead of lim(x->0) sin(x)/x which tends to 1 as well. At the end, the limit will tend to 2(x + 0/2) exp(x0 + 0²/2) = 2x.

[lucasfreire]

At 9:52 you can multiply top and bottom by 2xh + h² together with the fact that the limit as t approaches 0 of (a^t - 1)/t = ln(a) = 1 in this case since a = e. You'll have now e^(x²) times the limit as h approaches 0 of (e^(2xh + h²) - 1)/(2xh + h²) times the limit as h approaches 0 of (2xh + h²)/h. do t = 2xh + h² on the first limit and use the fact. The remaining limit will be (2xh + h²)/h as h approaches 0, this simplifies to 2x + h as h approaches 0 which is just 2x. Therefore you get e^(x²) * 2x as expected

[physicsatroeper]

I THINK you should be able to do the second problem directly with the limit definition of exp(u)=(1+x/n)^n as n --> infty. You have a double-limit, but I think it works out very cleanly unless I'm doing it wrong (quite possible: it's been years since I did these kinds of problems and I don't remember all the rules about when you're allowed to switch the order of the limits).

[stephenburrows]

Thanks for this, just beginning Limit Definition of Derivative in my AP Calculus class. Despite their lower level, they saw techniques that we have learned in the past few weeks. Trying to get them excited about this!

[seanfraser]

Dr. Penn's solution is completely valid if you define the exponential function by it's power series (which is standard in modern analysis). I have an alternative solution regardless.

We wish to evaluate the limit at 9:58. Put the e^(x^2) factor to the side for now and just focus on the quotient. Multiply the numerator and denominator by (2x+h). Then we have the limit as h->0 of (2x+h)(e^(2xh+h^2) - 1)/(2xh+h^2). Obviously 2x+h -> 2x as h->0. For the second factor, we use the well-known limit (e^u - 1)/u ->1 as u->0. I'll leave the proof of this fact as an exercise (how you prove it depends on how you define the exponential function, but none of the possibilities are difficult). Thus -> 2x as required.

[SlipperyTeeth]

9:52 You can do it using l'hopital. If the derivative exists at 0 or has a point discontinuity at 0, it follows that since the function is even, the derivative will approach 0 as the input approaches 0, which helps to simplify the final expression.

[Bodyknock]

11:23 "And we're left with h * (----) which tends toward zero as h goes to zero". I think this might be hand waving a bit, though, because the (----) is an infinite sum of terms and there's no explanation why you know (----) doesn't diverge toward infinity. The answer to that question would be that (----) is a factor of the Taylor expansion of eᵘ where u = 2xh + h², and since eᵘ is a finite number for any u that series must converge, and so therefore must its factors like (----). However, as Professor Penn said, that might be a "bit of a cheat" since the whole point of this exercise is finding a solution using the limit definition of the derivative, and it's not obvious that you can claim anything one way or another about a Taylor expansion converging without already having laid some groundwork of derivatives.

[knutthompson]

“There’s an easy way to do it, but let’s do it the hard way.” is the motto of calculus teachers everywhere. 😜

[MrRyanroberson]

9:50
L = lim h->0 of (e^(2xh+h^2) - 1)/h
L = lim h->0 of ((e^(2x+h))^h - 1)/h
as h approaches 0, if x is 0, then this is the limit of e^h^2
otherwise, h will be negligibly small compared to x.
first assume it is the latter case.
L(x > 0 or x < 0) = lim h->0 of ((e^(2x))^h - 1)/h
for this, as h goes to 0, this is the limit of the derivative of e^(2xy) with respect to y. the limit of such a derivative is ln(base) = 2x, QED b.
For the x=0 case, instead we must consider this:
L(x = 0) = lim h->0 of (e^h^2 - 1)/h
= lim h->0 of (e^h^2 - 1)/h^2 * h
= lim t->0 of (e^t - 1)/t * lim h->0 of h
= 1 * 0 = 0
this agrees with the (2x) derivative from the other case, so we conclude that this limit is always 2x.

[CM_France]

Hi,

1:02 : "plus" instead of "minus",

For fun:

1:21 : "ok, great",

10:13 : "and so on and so forth".

[uzytkownikgoogle]

7:15 Why is it correct to cancel out the h, despite having addition in the denominator?

[GaryFerrao]

9:40 it's not in the spirit of the video. the Taylor's Series expansion, uses derivatives, which by first principles uses limits. so it's kinda circular argument that you use this expansion to find the limit.

[sinecurve]

9:59 Using e := lim(h->0) (1+h)^(1/h) we have lim(h->0) ( (1+h)^(2x+h) -1))/h. Expand the first term in the numerator with a Bernoulli expansion. L = lim(h->0) (1 + (2x+h)*h + O(h^3) -1)/h = lim(h->0) 2x + O(h^2) = 2x.

Two points that I am wary on: (1) not sure if introducing e in this manner is kosher with the interchanging of limits and mixing of variables, and (2) 2x + h is not an integer so expanding this way probably involves Lambda functions to make this rigorous. Interested in anyone else's thoughts.

Edit: Bernoulli expansion of (a+b)^r is ok to use for all complex r.

[gniedu]

Since you're doing the derivative from the definition, shouldn't it be "illegal" to use Taylor series, due to the fact that Taylor series use information of the derivative?

[spicymickfool]

Might be worth noting the long way of finding the derivative of the square root function has the square root of x appearing on both side of the equation in the middle of the process. This allows re-representation of the square root as a continue faction. As it happens, the iterative process it implies for finding the square root yields essentially the same sequence as Newton's Method.