Math Cal1 final exam tomorrow. I went to "Organic tutor" and "Khan Academy" and didn't get this at all! I got their explanation, but they didn't deliver what I was looking for. (They just, for some reason, take the absolute value to be always positive. My prof stressed this wasn't the case in class, so I had to look for someone who could explain why) That's when I was like "Remember the guy who taught you how to find the derivative of the inverse functions so well? Well, look him up." And I did. I didn't remember the channel name so it took me a while to locate this video. And yes, it was worth it. you saved me. 😂. I immediately subbed. Your channel is so underrated. Thanks so much.❤
Thank you so much you really saved me. When you sit in a college class for 2 hours and the teacher does not stop writing they can't explain and thoroughly zoom in on the problem, like you do. Thank you so much, honestly you should have more subscribers.
Can I also use this if my problem is x2+x-16/ |x-8| if x approaches 8? Cause the way you explained it is way easier to understand than most of the videos I watched up until now
I think you should have used equals signs and not implications, since those aren't even statements. Atleast that's what i have been taught, where I live. Not sure if there are different conventions. I love the methodical and calm teaching style though.
Notice that u/|u| = sign u. Using this concept we have lim (X+3)(x-2)/|x-2| = lim (X+3 ) . Sign (x-2) = ±5 the sign depends on weather the limit is ↗️ or ↘️. Otherwise there is no limit.
![Limit of absolute value functions [ lim |x^3 - x|/(x^3 - |x|) as x goes to 0 ]](http://i.ytimg.com/vi/nJ9lEolfVpU/mqdefault.jpg)
Math Cal1 final exam tomorrow. I went to "Organic tutor" and "Khan Academy" and didn't get this at all! I got their explanation, but they didn't deliver what I was looking for. (They just, for some reason, take the absolute value to be always positive. My prof stressed this wasn't the case in class, so I had to look for someone who could explain why)
That's when I was like
"Remember the guy who taught you how to find the derivative of the inverse functions so well? Well, look him up."
And I did. I didn't remember the channel name so it took me a while to locate this video. And yes, it was worth it. you saved me. 😂. I immediately subbed. Your channel is so underrated.
Thanks so much.❤
Wow, I'm glad you found it helpful. Thanks for the sub, and please share with friends.
More hearts from Zambia, we can only say since we've learned we can testify that we're among the living......❤❤❤
Here I'm from India you make crystal clear to the confusion arises in such question very easily with broad smile. Thankyou
Happy to help!
Thank you so much you really saved me. When you sit in a college class for 2 hours and the teacher does not stop writing they can't explain and thoroughly zoom in on the problem, like you do. Thank you so much, honestly you should have more subscribers.
You make learning feel great. Second video in, I'm smiling from how much passion you put into your teaching!
Nicely done! And I love it that you are using chalk! No school like old school!❤
Thank you so much, I was so confused at first, but after a couple of minutes of watching this video, I feel MUCH better. Keep up the great work!
very good teaching sir. I love your teaching style.
This makes me understand clearly....keep it up sir
You are saving a lot sir thanks for your existence
best presentation keep on the good work my brother
As english is my second language you really helped me understand it thank you so much
Only Good video ive seen on this yet thanks so much saved my grade
Glad it helped!
Massive work done boss. I salute!
Can I also use this if my problem is x2+x-16/ |x-8| if x approaches 8? Cause the way you explained it is way easier to understand than most of the videos I watched up until now
Yes
You've made learning easy
But we can say that there is one limite from the rigt, and one from the left. So that the function is not diffenrtiable for 2
Thanks a lot dir 😊
I think you should have used equals signs and not implications, since those aren't even statements. Atleast that's what i have been taught, where I live. Not sure if there are different conventions. I love the methodical and calm teaching style though.
Thanks. Noted.
Very good. Thanks Sir
Superb teacher..
Great explanation
Prof is underrated😮😮❤
Thanks this helped me a lot
Great work Sir.
How do you manage to make it so simple 🤔
You are my hero
Notice that u/|u| = sign u. Using this concept we have lim (X+3)(x-2)/|x-2| = lim (X+3 ) . Sign (x-2) = ±5 the sign depends on weather the limit is ↗️ or ↘️. Otherwise there is no limit.
No, not in u = 0
🎉🎉🎉
Thank you
Thanks boss
Hello my friend you tesch nice
god bless !
thx
DNE
sharppppp
I am not the same now .... fire emojis
WolframAlpha says 5
R