14.2: Limits and Continuity

[@beyourself196]

your videos help us a lot. thank you from core of my heart

[@stnguyen1234]

Hey im taking Math 2415 right now. These videos are straight to the point and helpful!

[@ChrisHuxleys]

I had a question in my quiz yesterday that looks almost just like #6 and I couldn't figure it out for the life of me. THANK you for clearing it up for me! Your videos perfectly align to my Calc 3 class!

[@melchiortod29]

My teacher only brings limits where you can't plug it in😂 thank you very much! Well explained!

[@upsidedownChad]

Incrivel. Muito obrigado pela aula !!

[@danielrojas3958]

Nawebona me salvaste el primer parcial mi reina, excelente video, me suscribo

[@Salamanca-joro]

21:10 in this example i made y=0 and i got 1 for y axis and 1 for x axis as well so the limit does exist and it value is 1 right? Or we are supposed to prove that the limit does not exist?

[@alexandraniedden5337]

Showing that limits along two paths give the same value does not prove the limit is equal to that value. Oftentimes if we cannot substitute in the point, then the limit does not exist. The task then is to find two paths which give different limits; hence the limit would not exist.

[@Salamanca-joro]

@@alexandraniedden5337 oh yeah I forgot this! Thank you so much

[@ekindinek]

for the example 6 , 31:09 ; why did you only check the r goes to 0+ but not both 0+ and 0-

[@ekindinek]

i think it may because of lnx does not exist for negative side, but what if it would exist, shouldn't we check both sides?

[@alexandraniedden5337]

In polar, we take r to be positive, so we do not need to check 0- because this would be a negative value for r.

[@daddyji3742]

btw you are very wonderful teacher and love from India 🇮🇳

[@ChefFarisMom3]

I have a question for last questions. Sin theta Ln r applying LHP rule then it will be Cos theta .1/r?

[@HamzaAhmed-oq5od]

In the last example how do we get a 2r^2 and then the further 1/r in the numerator?

[@isisyasmim639]

At 33:18 I'm having a hard time figuring out where that 1/r and -1/2r² came from. Can you clarify, please?

[@alexandraniedden5337]

When you substitute r = 0 into the limit, you end up with infinity / infinity. This is a case where L'Hopital's Rule applies. After that infinity / infinity, I took the derivative with respect to r. The derivative of lnr is 1/r and the derivative of 1/(2r) is (1/2)*-1(r^-2). I hope this helps!

[@isisyasmim639]

@@alexandraniedden5337 Thank you! Now I get it!

[@justintang9801]

pretty sure at 13mins, the value of delta and absol. on the graph should be swapped with +/- delta at the x axis and absol. at the y axis, no?

[@elmao4927]

are there rules to the paths you can take? as in can you add constants so that y= x+ 5 for example

[@enriquesoler3150]

I have a question for these types of limits: after I try direct substitution and got an indeterminate form, do I just go straight into proving the limit DNE or first try limit rules first to see if I can get a limit value, and if that fails I prove the limit DNE?

[@alexandraniedden5337]

First try limit rules. Ultimately if you get an indeterminate form, the limit most likely doesn't exist, but not always.

[@vybsdeadly5493]

for the last example the limit as r --> 0 from the right is -infinite/ infinite ? becasue as ln goes to 0 y is going down forever ? so this confuses me because i didnt know lohopitals rule was applies this way

[@alexandraniedden5337]

Yes, you are correct! Should be negative infinity / infinity. And yes, you can use L'Hopital's Rule this way. Good catch!

[@daddyji3742]

are you teaching from thomas finney calculus mam ?

[@punyisasrimuang7782]

31:25 why y become r sin?

[@alexandraniedden5337]

I converted from rectangular to polar coordinates. For the conversion, x = rcos(theta) and y = rsin(theta). It will be easier to evaluate this limit using polar coordinates rather than rectangular.

[@prriyka]

You are incredible!

[@sfxyz3746]

You are really the best

[@苏伏]

any 14.3 video?

[@alexandraniedden5337]

No, sorry! Here is a video from a colleague covering 14.3 (his says 13.3, but it's 14.3 from the textbook I use - Partial Derivatives): kzbin.info/www/bejne/m4napmZ6oM9ob5Y

[@ERKBOI]

I cannot understand the last example. Can you clarify it a little bit? Where does rsin came from?

[@alexandraniedden5337]

The rsin(theta) is the polar conversion for y (think unit circle). Likewise, rcos(theta) is the polar conversion for x. Hope this helps!

[@hafiziehamdie8502]

have you covered chapter 14.3? or maybe it is on a separate video?

[@jkk23-g7c]

Great course, thank you very much. I wonder what textbook you are using. Is it the Stewart's Calculus textbook?

[@rayalhar]

Yes

[@olidialunga8593]

Did I miss where you ended up explaining "the other trick"

[@alexandraniedden5337]

Can you please give me a timestamp? I'm not sure what you are referring to.

[@olidialunga8593]

@@alexandraniedden5337 16:37, when mentioning the 3 options, option number 3 was 'one other "trick" ', which you had said you'd explain later. I didn't get that trick in the end. Did you mention it in this video?

[@alexandraniedden5337]

@@olidialunga8593 Ah, thank you. The "trick" I was referring to is finding the limit using polar coordinates rather than rectangular. See example #6 for this method.

[@olidialunga8593]

@@alexandraniedden5337 Oh, that makes sense. Thank you!

[@jkk23-g7c]

I had the same question, so thank you for asking this.

[@Salamanca-joro]

8:21 you could have just said : Since 0 does not equal -1/2 the limit does not exist at ( 0 , 0 ) right🤨?

[@alexandraniedden5337]

Yes, that is the ultimate answer to the problem :)

[@abeimnida]

I still dont understand why in example we need to consider the line y=x?

[@alexandraniedden5337]

I'm assuming you're referring to example 2. If so, the line y=x shows us that the limit as (x,y) approaches (0,0) is not the same along every path. I thought to choose y=x because that would give me -x^2 in the numerator, which I could eventually divide out with the denominator (not giving 0 as the previous 2 examples did). We had a feeling the limit would not exist so it was important to find a path where the limit did not = 0.

[@mukisajoshua2026]

I love youuuuuuu ❤️❤️❤️❤️ u saved my day

[@codingwithelhacen990]

21:39 . why did you chose y=x? can you also do limit along x-axis and y-axis?

[@alexandraniedden5337]

For example #4, I first approached the limit along the y-axis (which is x=0) - this was part a. I got the limit to be 1. If I had chosen to approach the limit along the x-axis (y=0), I would have also gotten 1. I could certainly have chosen to approach along the x-axis rather than the y-axis. Doing both isn't helpful (see next paragraph). In the case where a limit does not exist (which was likely the case here because we couldn't immediately substitute (0,0)), we need to find 2 differing limits. That's why I chosen y=x, because approaching along this path produces a limit of 0 (part b). The two differing values in parts a and b show that the limit does not exist.

[@abdullahalgarea3260]

@@alexandraniedden5337 Hello I saw some videos and I thing you are the best you tuber I have seen and I want to ask you if you have section 14.3

[@abdullahalgarea3260]

and I want to say to you keep going

[@alexandraniedden5337]

@@abdullahalgarea3260 I don't, but this is a video on partial derivatives (14.3) from a colleague of mine: kzbin.info/www/bejne/m4napmZ6oM9ob5Y Hope this helps!

[@benjaminandrew7940]

@@alexandraniedden5337 Thank you so much

[@HamzaAhmed-oq5od]

Are you sure the derivative for sinx lnx is correct?

[@alexandraniedden5337]

I am not sure where in the video you are referring, as I do not see an example with sinx lnx. That being said, the correct derivative is: sin(x)/x + ln(x)cos(x).

[@benjaminandrew7940]

is there a 14.3??

[@benjaminandrew7940]

I saw your reply to another person and see that you do not

[@govindc7690]

Can you please upload the notes?

[@kaneeindall5733]

Thank you!

[@aryankundu4954]

16:36

[@woldie2127]

great thanks

[@mukisajoshua2026]

Brilliant

[@abhijitkashyap3338]

Madam please solve the following f(x, y) = {1/x^2+y^2 (x, y) not eq (0, 0) 0 (x, y) = (0, 0)

[@ryanciminski4695]

Not the point of this channel.