L'Hospital's Rule - Indeterminate Products

[xfactor1223]

ur the only reason i have an A in calc 1 at johns hopkins

[patrickJMT]

@xfactor1223 spread the word about the videos : )

[sulmazz]

@xfactor1223 wow you must be really smart :-) good job !

[salnaser]

Thanks a lot! very helpful

[TheKingstoncowboy]

You've solved the equation to my heart......

[Itxturemom]

Hey I have a problem I'm stuck on the problem is Lim x-0 of xsin(1/x)

[sonu6797]

Your Videos are just great . I always study for exam from your videos .You should teach calculus at Rutgers :)

[patrickJMT]

@sonu6797 i have heard that before, actually : )

[Transane]

@sonu6797 as a fellow rutgers student, I agree. My current calc professor sucks. I dont bother showing up to class; simply watch your videos for half an hour on a bi weekly basis and I do a lot better then most everyone else.

[heybrian6]

for cotx, isnt the limit for as x approaches 0 negative infinity and positive infinity because on the open interval 0 to pi it approaches positive infinity but on open interval - pi to 0 it approaches negative infinity?

[Az094374]

2:40

Did you skip the chain rule in deriving x^(3/2) for a reason or was it just a mistake?

It really doesn't matter because it still becomes -3 * 0, and you get the same answer anyway. I'm just curious.

[UAEX110]

This is much too complicated for me T.T

[crazyasianskills]

If you have lim x->infinity of xtanx can you bring the tan under as tanx^-1 which is the same as inverse tan and take the derivative of that? or do you have to put it as cot?

[KingRobbStark]

@crazyasianskills 1/tanx is not the same as inverse tanx. 

[JewJitsu123]

@KingRobbStark It is. 1/tan(x) = tan^-1(x) = arctan(x)

[KingRobbStark]

@JewJitsu123 No. 1/tanx is cotx. Why? Because 1/tanx = 1/(sinx/cosx) and if you multiply by the reciprocal you should get cotx.

[patrickJMT]

@KingRobbStark you are correct of course

[patrickJMT]

@JewJitsu123 no, this is wrong. a super common mistake.

[victoraca]

thanks a lot

[victoraca]

Hei Patrick! Loving your videos but i´m having a problem here.

You say 1:35 that

ln x / x^-1/2 is in the form: infinitive over infinitive but it sounds like  -infinitive/0 to me

help me out.

[patrickJMT]

@victoraca well in the denominator you have x^(-1/2) which is 1/x^(1/2)

as x --> 0 , 1/x^(1/2) goes to infinity

[victoraca]

@patrickJMT thanks a lot

[victoraca]

@patrickJMT thanks a lot

i´m a little stupid at basics math...or just math

[sullivanseven]

Hi Pat, at 1:35 you say that x^-1/2 is infinity while I think it's zero because 0^-1/2 is zero.

Am I right?

[andreiduffy]

THANK YOU SO MUCH

[Catergory5Hurricane]

These limits can also be found in power series.

[Skynt]

I'm having terrible trouble with this limit problem:

lim x->0  (2e^(-1/x^2))/x^3

I try L'H but I usually get like 0 over infinity which I don't think I can conclude is 0 ...

[patrickJMT]

i dunnooo.... zero divided by a large number sounds like it would be close to zero to me....

[Cammie010]

@Skynt @Skynt Hi there.

I know it's been over a year but I'm bored and your problem interested me. (I must be really really bored..)

After applying L'Hopital's (Bernoulli's :p ) rule, i got lim x-> 0 (-3x^-1)/(e^(x^-2)).

the function is between -3/2x and 3/2x in the open interval (-1,1). Since -3/2x and 3/2x both approach 0 as x->0, by the Squeeze Thereom the original limit also approaches 0.

I'm pretty sure this is correct but PLEASE let me know if you find errors. thanks :D

[Cammie010]

@Cammie010 btw, apply L'Hopitals twice (the function I mentioned above wasn't the function to use the squeeze theorem) to get 3/2(x/(e^(x^-2)).

[JustSemantics]

I think supergsx is wondering what the difference is between limits approaching different sides of a point.

Great videos by the way, my school unfortunately doesn't have any calculus classes, so this is the best way to learn it!

[supergsx]

how do you tell the difference between the limit of a+ and a-?

[patrickJMT]

i am not sure i understand your question

[supergsx]

...k so youre putting limit from x to 0+ ...

How do you find the + answer different from the - answer?!?!!

[patrickJMT]

the function is not defined for values of x that are less than or equal to 0; therefore, taking the limit from the left of zero would not make sense to do

[supergsx]

ok i get that but i'm not exactly talking about your video.

It's a simple math question.

How, using algebra, can you find a limit from the left as opposed to a limit of the same number from the right?

[patrickJMT]

you do it the same way... you just have to account for signs in most of the problems