Thanks you so much! I don't have to take Calc 3 at my university, but I've always wanted to learn it. Your videos have been so helpful for me and I really appreciate it!
Another possibility for #2: Consider the denominator (i.e. a^2-b^2) as the result of (a+b)*(a-b) (plus-minus binomial formula). (a-b) is given in the numerator as well, thus both cancel each other. 1/(a+b) remains or the denominator equals both sqare roots added.
At 18:08 why can’t you just substitute r=0 into the bottom then leave it as (rcos^3@)/cos^2@ then cancel out the cos to get rcos@ then substitute r=0 again to get the limit as 0?
@@krithikkumar9887 two things. Doesn’t cos^2(@)/cos^2(@) approach 1 as @ approaches pi/2 anyway? Also as @ is an introduced variable, I thought we were allowed to cancel?
My way: set x=y, then use the l'Hopital's rule. It is probably not mathematically correct, but it works, if both x and y are aproaching the same value.
Just a problem: there's a possibility that the limit does not exist. The issue with multivariable derivatives is that there are infinite directional limits and not just left and right sided. It is possible that limit x=y is not the same as lim x=0 or y=0 etc
You can use L’Hopital’s rule and you’ll get the same answer. The derivative of sin(t) is cos(t) and the derivative of t is 1. Plug in 0 to t and you’ll see that cos(0) is 1.
Hi man, I can explain to you. The reason is that the limit of sin(h)/h as h goes to 0 appears on the derivative of sin(x) by definition. So, to know that the derivative of sin(x) is cos(x), you need to solve that limit first, resulting in a cyclic thing, get it?
@@griffinf8469 yeah for sure, but what if you were to do the derivative of sin(t) using the limit definition? You would find this limit while working it out, and then you'll be tempted to use L'Hopital's rule. But that just doesn't make sense because you are taking tyhe derivative of sin(t) in the first place. That's why you have to find other ways to do it, preferably the squeeze/sandwich theorem
How to show nonexistence: ua-cam.com/video/yGPsofTdtQw/v-deo.htmlsi=iPH8WT3GM-nsgFcA
Please, I want Calculus3 manual.
Thanks you so much! I don't have to take Calc 3 at my university, but I've always wanted to learn it. Your videos have been so helpful for me and I really appreciate it!
Take it for fun
The polar blooper lol
I laughed out loud at 15:30 when you drew the frowny face saying that there was no conclusion. I don’t know what it was, but it just got me.
forgot to edit the middle part😂😂
@@Vernonnotyourman It's basically becoming a feature in his channels
Be careful when we use the polar method:
ua-cam.com/video/gL2fzELFFwQ/v-deo.html
Thank for not editing that
I appreciate that
(Not sarcastic)
Another possibility for #2: Consider the denominator (i.e. a^2-b^2) as the result of (a+b)*(a-b) (plus-minus binomial formula). (a-b) is given in the numerator as well, thus both cancel each other. 1/(a+b) remains or the denominator equals both sqare roots added.
Thank so much 🎉
At 18:08 why can’t you just substitute r=0 into the bottom then leave it as (rcos^3@)/cos^2@ then cancel out the cos to get rcos@ then substitute r=0 again to get the limit as 0?
Again what if theta is pi/2, then u can't cancel cos
@@krithikkumar9887 two things. Doesn’t cos^2(@)/cos^2(@) approach 1 as @ approaches pi/2 anyway? Also as @ is an introduced variable, I thought we were allowed to cancel?
Thank you very much! Wonderful
L'Hopital's rule is haunting this man.
I still have nightmares with Epsilon-delta proofs in multivariable calculus
Can someone explain me why (e^2x - 1)/x is equal to f'(0)? ( I only know math up to derivatives)
one definition of derivatives is lim x->x_0 (f(x)-f(x_0))/(x-x_0). Plugging is 0 for e^2x we get: lim x->0 f(x)-f(0)/x-0 which is thr expression above
@@t.b.4923 I see now,thank you
Is there any way to tell whether you should start using the path method to show the limit does not exist or start the Squeeze Theorem?
My way: set x=y, then use the l'Hopital's rule. It is probably not mathematically correct, but it works, if both x and y are aproaching the same value.
Just a problem: there's a possibility that the limit does not exist. The issue with multivariable derivatives is that there are infinite directional limits and not just left and right sided. It is possible that limit x=y is not the same as lim x=0 or y=0 etc
Has there been a video showing where the limit of a product exists, but the product of the limits does not exist?
You can just use 1/x and x as x approaches 0.
Is there an equivalent theorem of L'Hopital rule in calc 3?
Unfortunately no
@@anthonyflanders1347 shucks!
Thanks for the answer tho :)
How are you getting the search option in comments?
8:18 another new tool learnt today
Multivarible function is multiterrible
Multiterrible calculus 😔
Now number 5, the larch.... the larch
He wrote squezze lol
Why can’t L’Hopital’s rule be used for the substitution method? It’s single-variable?
You can use L’Hopital’s rule and you’ll get the same answer. The derivative of sin(t) is cos(t) and the derivative of t is 1. Plug in 0 to t and you’ll see that cos(0) is 1.
@@griffinf8469 yeah, that’s how I would approach it. Just curious as to why bprp said you couldn’t use it
@@chonkycat123he doesn't like using it for some reason, I've noticed it in other videos as well
Hi man, I can explain to you. The reason is that the limit of sin(h)/h as h goes to 0 appears on the derivative of sin(x) by definition. So, to know that the derivative of sin(x) is cos(x), you need to solve that limit first, resulting in a cyclic thing, get it?
@@griffinf8469 yeah for sure, but what if you were to do the derivative of sin(t) using the limit definition? You would find this limit while working it out, and then you'll be tempted to use L'Hopital's rule. But that just doesn't make sense because you are taking tyhe derivative of sin(t) in the first place. That's why you have to find other ways to do it, preferably the squeeze/sandwich theorem
😊 👍
First