Awesome!! Videos will be posted throughout the year- AP Calc AB is the course I teach 🤗 Please help spread the word to others who might find this channel helpful!!
This is truly a life saver! I was in total despair in my calculus course thinking it is just IMPOSSIBLE, but now I see it's not that bad thanks to you. Thank you zilions!!
Im so happy to hear that. My school year starts next week and then you can expect regular content throughout the whole year. If you have trouble with anything let me know. And help me spread the word about about the channel! We are kind of new and want to help as many kids as possible! We feel so bad for what you guys are going through! Have a great day!
This is really, really good. I want to get a fast as you are with the algebra. Can you recommend any videos. I love your approach and clarity. I am specifically interested in fractions and factoring. Many thanks!
... A good day to you, When I finished watching evaluating the indeterminate limit #5, I was thinking of an alternative way to solve the same limit as follows: lim(x-->4)((sqrt(x + 5) - 3)/(x - 4)) (Rewrite the denominator (x - 4) as: (x - 4) = (x + 5) - 9 and treat this new expression as a difference of two squares: (x - 4) = (x + 5) - 9 = (sqrt(x + 5) - 3)(sqrt(x + 5) + 3) to finally cancel the common factor (sqrt(x + 5) - 3) of numerator and denominator to obtain the solvable limit form: lim(x-->4)(1/(sqrt(x + 5) + 3)) = 1/(3 + 3) = 1/6 ... to my surprise it works out; hope you appreciate this alternative approach too ... Thank you for your great math efforts, Jan-W
Whoa whoa whoa 4:25 this limit is NOT DNE! Do not write that people! Just because the function does not exist where x is approaching, does not mean there isn’t a limit. It means there is either a vertical asymptote, jump, or a removal discontinuity (hole) in your graph. If you get a number/0 after plugging in, that means that your answer can be DNE, infinity, or negative infinity.
Hi TheCatch- unfortunately you are incorrect. We aren’t looking at one sided limits here, if we were we could describe the limit as positive or negative infinity. The definition of a limit specifically states that the answer must be a number and infinity is not a number. If there was a removable discontinuity we would have gotten 0/0 when originally plugging in. There is most definitely not a jump discontinuity since this is not a piecewise function and is comprised of a quotient of polynomials (which is the type of functions focused on in this video). Since we got a number over a 0, that alludes to a vertical asymptote, which means the graph is approaching + or - infinity which again, cannot be stated as answer to this problem. We say the limit does not exist, and would have to do further investigation using one sided limits to accurately describe the behavior of the function. Hope this helps your confusion. 😊
@@themathandphysicstutor looks like I thought that the limit was as x approached 0, not 2. That was my problem. If it was approaching 0 it would be 1/2.
Can’t we just find the limit as x-> 0 by finding the limit as x-> a number really close to 0 like 0.0001. By doing this, we can technically figure out where y is going as we get closer and closer to 0 so we can most likely find the limit. But I’m not sure. I’m just stating what I thought. Could be wrong😅
Good work, could I say when a function itself is not able to produce output at a specific input and it haults the operation at a specific x value. To overcome this situation we take help from the limit and limit helps us to take out of this situation. After calculating the limit value we GIVE THIS VALUE TO THAT FUNCTION WHICH WAS Not PRODUCING ANY OUTPUT AT SPECIFIC INPUT??? iF THIS IS THE CASE THEN can say that function and limit both are different BUT it is the limit which helps the function which was trapped on a specific value and now because of limit function is able to walk smoothly again?? In a result could I say that whenever a function trapped in an undefined situation, we have to bring in LIMIT at this point?? please help me out! thanks.
How do I solve if its like this lim { 2, x< -1 { x^2-2x+1, x>= -1 x-> -1 both of the limit is different, so does this mean the limit does not exist? (I got 2, and for the bottom I got 4)
In seven minutes you just taught me what my calculus teacher couldn’t in two weeks. Thank you!
Glad to help
I just bombed a test over this exact subject and now I see it was so easy 😂
Less, the intro took a minute
Could not have been clearer! You just cured me of my looming AP calc stress entirely, what an absolute life saver.
Thank you so much for the kind words! Glad I could help! Please share with your friends
Oh my gosh! It's like you are a mind reader with topics! I just am starting AP Calc AB!!!
Awesome!! Videos will be posted throughout the year- AP Calc AB is the course I teach 🤗 Please help spread the word to others who might find this channel helpful!!
My calculus revision was completed by this ... ❤️ Thanks from india ...
Thank you
Such an underrated channel you just nailed it ma'am😊
Thank you so much 😀
This is truly a life saver! I was in total despair in my calculus course thinking it is just IMPOSSIBLE, but now I see it's not that bad thanks to you. Thank you zilions!!
You're very welcome!
just started ap calc and this helped a lot
Im so happy to hear that. My school year starts next week and then you can expect regular content throughout the whole year. If you have trouble with anything let me know. And help me spread the word about about the channel! We are kind of new and want to help as many kids as possible! We feel so bad for what you guys are going through! Have a great day!
This is really, really good. I want to get a fast as you are with the algebra. Can you recommend any videos. I love your approach and clarity. I am specifically interested in fractions and factoring. Many thanks!
Thank you so much. It comes with practice
... A good day to you, When I finished watching evaluating the indeterminate limit #5, I was thinking of an alternative way to solve the same limit as follows: lim(x-->4)((sqrt(x + 5) - 3)/(x - 4)) (Rewrite the denominator (x - 4) as: (x - 4) = (x + 5) - 9 and treat this new expression as a difference of two squares: (x - 4) = (x + 5) - 9 = (sqrt(x + 5) - 3)(sqrt(x + 5) + 3) to finally cancel the common factor (sqrt(x + 5) - 3) of numerator and denominator to obtain the solvable limit form: lim(x-->4)(1/(sqrt(x + 5) + 3)) = 1/(3 + 3) = 1/6 ... to my surprise it works out; hope you appreciate this alternative approach too ... Thank you for your great math efforts, Jan-W
Whoa whoa whoa 4:25 this limit is NOT DNE! Do not write that people! Just because the function does not exist where x is approaching, does not mean there isn’t a limit. It means there is either a vertical asymptote, jump, or a removal discontinuity (hole) in your graph. If you get a number/0 after plugging in, that means that your answer can be DNE, infinity, or negative infinity.
@@themathandphysicstutor I’m right arent i?
Hi TheCatch- unfortunately you are incorrect. We aren’t looking at one sided limits here, if we were we could describe the limit as positive or negative infinity. The definition of a limit specifically states that the answer must be a number and infinity is not a number. If there was a removable discontinuity we would have gotten 0/0 when originally plugging in. There is most definitely not a jump discontinuity since this is not a piecewise function and is comprised of a quotient of polynomials (which is the type of functions focused on in this video). Since we got a number over a 0, that alludes to a vertical asymptote, which means the graph is approaching + or - infinity which again, cannot be stated as answer to this problem. We say the limit does not exist, and would have to do further investigation using one sided limits to accurately describe the behavior of the function. Hope this helps your confusion. 😊
@@themathandphysicstutor looks like I thought that the limit was as x approached 0, not 2. That was my problem. If it was approaching 0 it would be 1/2.
Is the solution for lim x approaches to -2 |x+2|/(x+2) equal to does not exist or i can do something more?
For problem 3, L Hopitals Rule can be applied. Apply the derivative to the numerator and denominator. Plug in 2 for x. The answer is 12.
You're saving my grade!
That’s what I’m here for. Please share with anyone else who needs savings
the method in number 2 is the difference of two squares, right? I'm looking forward to your response. Thank you!
The denominator uses difference of two squares. The numerator uses sum of cubes. :)
thanks for your help, last question on assignment was evaluating limits by hand. normally we just use maple. I got the question right
Excellent!
if you arrive at #/0, this does not necesarily mean that the limit is DNE. I am seing many cases in which they do in fact exist, while being #/0.
Sorry, so ive figure out, if x --> 0, its in these cases where #/0 is not DNE
Can’t we just find the limit as x-> 0 by finding the limit as x-> a number really close to 0 like 0.0001. By doing this, we can technically figure out where y is going as we get closer and closer to 0 so we can most likely find the limit. But I’m not sure. I’m just stating what I thought. Could be wrong😅
Thank you for posting this. Very concise and helpful
You just saved my grade 🥳 This is really great 💓
Glad I could help!
This was really helpful thank you! Perfect video!!
You’re welcome
Good work, could I say when a function itself is not able to produce output at a specific input and it haults the operation at a specific x value. To overcome this situation we take help from the limit and limit helps us to take out of this situation. After calculating the limit value we GIVE THIS VALUE TO THAT FUNCTION WHICH WAS Not PRODUCING ANY OUTPUT AT SPECIFIC INPUT??? iF THIS IS THE CASE THEN can say that function and limit both are different BUT it is the limit which helps the function which was trapped on a specific value and now because of limit function is able to walk smoothly again?? In a result could I say that whenever a function trapped in an undefined situation, we have to bring in LIMIT at this point?? please help me out! thanks.
0:27 isn't 4th exactly like 1st? In both situations you get a direct answer. It's just a whole number.
Many students get confused with an undefined answer vs. a 0 answer, hence the distinction.
Very well organized and explained! 😊
Thank you! 😊
super helpful, thank you!
Thank you
Example number 2 isn't it supposed to be 1, because -2² is -4 which will make the numerators -4+4+4=4 and the denominator =4 which will be 4/4=1?
it's actually (-2)^2, which evaluates to 4.
This video could've save me from my first quiz disaster lmao
thank you so much it helps
Wow brilliant work there
Thank you
How do I solve if its like this
lim { 2, x< -1
{ x^2-2x+1, x>= -1
x-> -1
both of the limit is different, so does this mean the limit does not exist?
(I got 2, and for the bottom I got 4)
What a great use of 7 min!
I agree
dam you really explained everything so clearly !
Thank you
Thank you for this tutorial
You're welcome 😊
where did x-4 come from for the top on number 5? i thought we had x-5+9 up there
-5+9=4 :) we just combined
@@themathandphysicstutor thank you! 🤦♂️
Thank you this helped me
No problem
Thank you so much!
You're welcome!
good lecture thank you.
You’re welcome
thank you ☺
You're welcome 😊
Amazing.
Thank You!!!♥👍
You are so welcome
Marvelous love from Pakistan ❤
Thank you
awesome!
Thank you! Cheers!
You should really set up a Patreon account for donations.
Thank you. You can always leave a “super thanks”. We really are just here to help! Thanks again for the consideration
❤
why did she forget the negatives
Where?
I just want to ask questions so can you please send me your email
My tutoring schedule is currently booked. I’ll let you know if I have cancelations. Usually I book up mid school year. Thank you