so you want to use L'Hospital's Rule?

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2 жыл бұрын

So you want to use L'Hospital's Rule? Then use it carefully!
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Пікірлер: 71

@pneujai 2 жыл бұрын

calculate 😔 evaluate 🤗

@bprpcalculusbasics 2 жыл бұрын

😆

@neilgerace355 2 жыл бұрын

L'Hospital's Rule is great but yes we always have to think about its limitations.

@tubax926 2 жыл бұрын

Yup, that's when you use L'Hopitals rule, its big brother.

@joakimharbak7485 2 жыл бұрын

Don't you mean; its limits?

@78anurag Жыл бұрын

@@joakimharbak7485 Lmao thought the same

@gammano0b858 2 жыл бұрын

I did it by replacing x with 1/t, so I ended up with the limit of e^-t * t or t / e^t. Then notice that the exponential function grows faster than any power, so its pretty clear that its going to zero, ptherwise L'Hopital would do the job here too :) Great Videos from you, always nice to see those traps and how to avoid them.

@mangeshhebbalkar1715 2 жыл бұрын

even i did that

@createyourownfuture3840 2 жыл бұрын

L'hôpital's rule does that in the background actually...

@chrisrybak4961 2 жыл бұрын

Nice, neat demonstration of how to choose how to apply L’Hôpital’s rule! Fun that as x -> 0- this function goes to -infinity…

@hassanalihusseini1717 2 жыл бұрын

Yes, a nice trick to apply the Hospital rule! Thanks for showing.

@user-wu8yq1rb9t 2 жыл бұрын

Great Thank you so much *Dear Teacher* 💖

@user-wu8yq1rb9t 2 жыл бұрын

You're so smart teacher! You know when you should upload your new videos (beautiful videos).

@YU-zg7zg Жыл бұрын

This has helped so much right before my exam! I almost gave up trying to figure out a similar problem. Thank You!

@gustavocortico1681 2 жыл бұрын

It's fun to see that the power rule in the first situation dominates the expression after an arbitrary number of l'h rules, making the evaluation of any positive x closer and closer to zero

@rshawty 2 жыл бұрын

you can also see it from the the first method, by contradiction Suppose L = lim [( e^(-1/x) )/x], x->0+ is finite, then we see it’s a 0/0 situation so indeed we can use L’Hopital’s rule : L = lim [( 1/x² • e^(-1/x) )/1], x->0+ L = lim [( e^(-1/x) )/x²], x->0+ L = (lim [( e^(-1/x) )/x], x->0+)(lim [1/x], x->0+) L = L•∞ = ∞ So if L converges it implies L diverges, which is impossible. Hence L diverges.

@moskthinks9801 2 жыл бұрын

Except that L can be 0, where 0 = 0•∞, although the proof is quite interesting

@rshawty 2 жыл бұрын

@@moskthinks9801 ahh yes oups

@lngbrmchryshorts9748 Жыл бұрын

Thank you sir

@2gr_t95 2 жыл бұрын

i'm in first year of french ingeneering preparatory class but as i'm in a course more focused on chemistry, L'hospital's rule isn't being taught to us we're just given classic equivalents of functions and it works just fine :,)

@Reluxthelegend 2 жыл бұрын

"the sad face is never the answer"

@emilsriram92 2 жыл бұрын

lmao, i just had this question on my homework assignment on my calc course.

@robertveith6383 2 жыл бұрын

*@blackpenredpen* -- Near the end where you cancel the -1/(x^2) from numerator to denominator, it would be better to put grouping symbols around the -1/(x^2) for clarification and emphasis, as it is being multiplied by that other term in the denominator.

@GhostHawk272 2 жыл бұрын

There is a dot

@ytsimontng 2 ай бұрын

It would be better to call it "that other factor" instead of "that other term" for clarification and emphasis, as it is being multiplied.

@domanicmarcus2176 2 жыл бұрын

Wil the left-hand limit equal the right-hand limit? Will the overall limit equal to zero or is the left-hand limit not equal to the right-hand limit?

@FreshBeatles 2 жыл бұрын

Hospital rule!

@seanmurphy2278 2 жыл бұрын

In the first attempt using L'Hopitals rule increases the exponent below the line hence decreasing the value of the function. Can we say that this proves that the function goes to 0 or does x=0 make the increasing exponent redundant.

@Bruh-bk6yo 3 ай бұрын

Ugh... ln(e^(-1/x)/x)=-1/x+ln(1/x) but ln(1/x) is always smaller than 1/x for x>0, therefore we get: -1/x+o(1/x) for each x>0 there will be an eps from (0;1) such that ln(1/x)=eps/x. Then, our limit is the same as the limit -(1-eps)/x for x -> 0. 0 ln(...) -> -∞, which means that the given limit equals 0.

@iDovahkiin 2 жыл бұрын

Could someone explain why when he took e^(-1/x) to the denominator nstead of 1/x•e(^1/x) He wrote 1/x/e^(1/x)????

@asteriskconfidential7403 2 жыл бұрын

Could use brute force(Taylor series) instead of hospital rule

@alejandrojara1957 10 ай бұрын

When you derivate first time, left side is correct. You just have take limit approach zero, and e to the power - infinite is zero.

@ianthebadguy 2 жыл бұрын

I just noticed your bulk boxes of Expo markers... do you think they offer endorsement deals? :-D

@bprpcalculusbasics 2 жыл бұрын

Not sure. I bought them myself tho.

@ianthebadguy 2 жыл бұрын

@@bprpcalculusbasics I'm only half-joking... it might just be in their interest that a prolific math KZfaqr continues to be seen using Expo markers instead of the generic brand, lol

@schizoframia4874 2 жыл бұрын

Using lhopitals rule for |x|/x as x goes to zero is a big middle finger

@givrally7634 2 жыл бұрын

Doesn't L'Hopital's rule only work for analytical functions ?

@reeeeeplease1178 2 жыл бұрын

Well they need to be differentiable at the limiting x value

@pauljackson3491 2 жыл бұрын

Let y = e^(-1/x)/x and z = e^(-1/x)/x^2 Using L'hopital's rule with y(x) just gets us z(x) which is worse. But since lim(y) = lim(z) that means lim(y)/lim(z) = 1 which means lim(y/z) = 1 Which isn't true. Does the indeterminateness mean I can't multiply them or combine the limits? And if so is there a way of doing something like that to find limits using l'Hopital rule?

@NoNameAtAll2 2 жыл бұрын

lim(y) = lim(z) => lim(y/z)=1 only works if it's not 0/0 or inf/inf

@oenrn 2 жыл бұрын

0/0 is indeterminate, meaning the limits can be anything and aren't necessarily the same. lim y = 0/0 and lim z = 0/0 does not imply lim y = lim z, as both 0/0 can have diferent values.

@Cobalt_Spirit 2 жыл бұрын

I have a question: what is the limit as x approaches 1, of the logarithm base x of 10?

@rshawty 2 жыл бұрын

So your question is : L = lim [log_x(10)], x->1. Rewrite the limit : L = lim [ (ln10)/(ln(x)) ], x->1 = (ln10)/ln(1) = (ln10)/0 = ln10 • ∞ = ∞.

@redwanburkan4790 2 жыл бұрын

Well, the limit here doesn't exist as long as we are approaching 1 from both sides And why it doesn't exist? That's simply because the base can't be 1 (from logarithms properties)....... It's just like when you try to find the limit of (1/x) when x approaches 0 (which is also doesn't exist)

@Cobalt_Spirit 2 жыл бұрын

@@rshawty Is that a positive or a negative infinity?

@rshawty 2 жыл бұрын

@@Cobalt_Spirit neither my bad, I shoulded have done the limit from the right and from the left and concluded that the limit does not exist for an "exact" 1 as @Redwan Burkan said

@rshawty 2 жыл бұрын

@@Cobalt_Spirit because the limit as x approaches 1+ is +inf, and the limit from the left is -inf

@bertrandviollet8293 2 жыл бұрын

You can guess 0 is the limit by seeing that the denominator is exploding each time you derive and repeat derivating

@reeeeeplease1178 2 жыл бұрын

Each iteration, the denominator gets smaller since we are approaching 0, so the whole limit would get bigger

@tarehjernetarehjerne4082 2 жыл бұрын

I don't recall it being called L'Ho*S*pital's rule If i was a teacher i would take points for that lol

@Cloud88Skywalker 2 жыл бұрын

I thought you were going to do L'H by integrating top and bottom!! I mean taking the -1st derivative! so you'd get lim (e^(-1/x) / ln(x)), as x->o+, you get 0/(-∞) = 0 But I guess it's not correct to do that :(

@p12psicop 2 жыл бұрын

Take the limit with everything in the denominator and no need for LH rule.

@bubbeldiamendo 2 жыл бұрын

I can't understand one thing. You can't use L'Hospital's Rule on limit _(x->0) (sinx/x) because of definition of derivative. So why you are using it here despite there is (.../x)?

@reeeeeplease1178 2 жыл бұрын

Because this isnt the definition of the derivative of a function (atleast not a "pure" function like e^x or smth) If you were to consider f(x) = e^(-x) and wanted to calculate f'(0), we would get a similar limit but not quite the same *BUT* mathematically speaking, you can use L'H whenever the limit allows it, even for sin(x)/x. BPRP argument against using L'H is that sin(x)/x needs to be calculated to get (sin(x))', so we have to act like we didnt know its derivative yet... Not really a fan of this since you can calculate (sin(x))' in other ways...

@eorojas 2 жыл бұрын

And limit for 0-?

@oenrn 2 жыл бұрын

Just do the same thing but replace the signs at the end. 1 / (e^(1/0-) = 1 / e^-inf = 1 / 0+ 》because exponential is always positive = + infinity

@oftenbryan 2 жыл бұрын

Isn't it told that exp(x) converges faster than x therefore we only need to care about the limit of e^-(1/x) --> 0 (at 0+)

@krabbediem 2 жыл бұрын

Why does L'Hopital's rule fail in the original case?

@Dalton1294 2 жыл бұрын

Editing to to say that the rule fails because the initial question creates a never ending cycle

@ffc1a28c7 2 жыл бұрын

@@Dalton1294 They're asking why it does. That is how it fails.

@krabbediem 2 жыл бұрын

Hi Dalton G, and thank you for answering. Yes, I think BPRP illustrated this point pretty well. I was more interested in if these cases had a type of identifer, so that I wouldn't just iterate away on L'Hopital without rewriting. Sorry I wasn't clearer on that. I see now that my question was poorly formulated.