so you want to use L'Hospital's Rule?
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- Опубліковано 7 кві 2022
- So you want to use L'Hospital's Rule? Then use it carefully!
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L'Hospital's Rule is great but yes we always have to think about its limitations.
Yup, that's when you use L'Hopitals rule, its big brother.
Don't you mean; its limits?
@@joakimharbak7485 Lmao thought the same
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.
even i did that
L'hôpital's rule does that in the background actually...
Nice, neat demonstration of how to choose how to apply L’Hôpital’s rule!
Fun that as x -> 0- this function goes to -infinity…
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
Yes, a nice trick to apply the Hospital rule! Thanks for showing.
This has helped so much right before my exam! I almost gave up trying to figure out a similar problem. Thank You!
Great
Thank you so much *Dear Teacher* 💖
You're so smart teacher!
You know when you should upload your new videos (beautiful videos).
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.
Except that L can be 0, where 0 = 0•∞, although the proof is quite interesting
@@moskthinks9801 ahh yes oups
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.
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 :,)
Thank you sir
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.
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?
*@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.
There is a dot
It would be better to call it "that other factor" instead of "that other term" for clarification and emphasis, as it is being multiplied.
lmao, i just had this question on my homework assignment on my calc course.
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)????
Hospital rule!
When you derivate first time, left side is correct. You just have take limit approach zero, and e to the power - infinite is zero.
"the sad face is never the answer"
Could use brute force(Taylor series) instead of hospital rule
Doesn't L'Hopital's rule only work for analytical functions ?
Well they need to be differentiable at the limiting x value
Using lhopitals rule for |x|/x as x goes to zero is a big middle finger
I just noticed your bulk boxes of Expo markers... do you think they offer endorsement deals? :-D
Not sure. I bought them myself tho.
@@bprpcalculusbasics I'm only half-joking... it might just be in their interest that a prolific math UA-camr continues to be seen using Expo markers instead of the generic brand, lol
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?
lim(y) = lim(z) => lim(y/z)=1 only works if it's not 0/0 or inf/inf
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.
I have a question: what is the limit as x approaches 1, of the logarithm base x of 10?
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 • ∞
= ∞.
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)
@@rshawty Is that a positive or a negative infinity?
@@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
@@Cobalt_Spirit because the limit as x approaches 1+ is +inf, and the limit from the left is -inf
You can guess 0 is the limit by seeing that the denominator is exploding each time you derive and repeat derivating
Each iteration, the denominator gets smaller since we are approaching 0, so the whole limit would get bigger
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
Take the limit with everything in the denominator and no need for LH rule.
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)?
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...
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 :(
And limit for 0-?
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
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+)
Why does L'Hopital's rule fail in the original case?
Editing to to say that the rule fails because the initial question creates a never ending cycle
@@Dalton1294 They're asking why it does. That is how it fails.
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.
Is it "hospital's rule" or "hopital's rule?"
Nobody notces that it is "hospital" instead of "hopital"
Isn’t it “L’Hopital”?
Yes... That is what i had heard in my classes
419 + 1 likes and 17 hours late moment
Lol nice troll with lhopital's rule
Also I like how everyone is calling it l'hospital rule lol
Hospital = Hôpital because
Ô = Os (historically)
In original H's books you'll find his name as Guillaume de l'Hospital
@@SimsHacks cool I didn't know that
Now it makes more sense
But it doesn't make it less funny lol
It's L'Hôpital, not a hospital