Thank you so much for this video, I was struggling with indeterminate form and this was the only thing that helped me understand so clearly. Keep up the good work!
... After watching your clear presentation I suddenly saw an alternative way to solve the same indeterminate limit: lim(x-->0)(x/(sqrt(x + 3) - sqrt(3)) , Rewrite numerator x as follows: x = (x + 3) - 3 and treat this new expression as a difference of two squares: x = (x + 3) - 3 = (sqrt(x + 3) - sqrt(3))(sqrt(x + 3) + sqrt(3)) to finally cancelling the common factor (sqrt(x + 3) - sqrt(3)) of numerator and denominator, obtaining the solvable limit form of: lim(x-->0)(sqrt(x + 3) + sqrt(3)) = 2sqrt(3) ... Hoping this way is also appreciated a bit ... Thank you for your great math efforts, Jan-W
Thank you so much for this video, I was struggling with indeterminate form and this was the only thing that helped me understand so clearly. Keep up the good work!
Thank you for this video. I was struggling with this concept and your explanation was easy to follow.
... After watching your clear presentation I suddenly saw an alternative way to solve the same indeterminate limit: lim(x-->0)(x/(sqrt(x + 3) - sqrt(3)) , Rewrite numerator x as follows: x = (x + 3) - 3 and treat this new expression as a difference of two squares: x = (x + 3) - 3 = (sqrt(x + 3) - sqrt(3))(sqrt(x + 3) + sqrt(3)) to finally cancelling the common factor (sqrt(x + 3) - sqrt(3)) of numerator and denominator, obtaining the solvable limit form of: lim(x-->0)(sqrt(x + 3) + sqrt(3)) = 2sqrt(3) ... Hoping this way is also appreciated a bit ... Thank you for your great math efforts, Jan-W
Helps a lot! Thank you so much sir!
Thank you for this sir😊 I learned something today😁
Thank you so fucking much I wish you the best in the world
he is good
It should be 5 (3-x) and not 5(3-x)
9:59 😊
No..
Aren’t they the same thing?
Dude you can just diffentiate by using
L hospital's rule and you will get the answer in two steps
wtf i dont understand. am I that noob?
yes