Finding Limits an Algebraic Approach
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- Опубліковано 10 лип 2024
- In this video we will find limits of functions algebraically using simplification methods such as factoring, rationalizing, and simplifying.
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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!!
... 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
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!
Is the solution for lim x approaches to -2 |x+2|/(x+2) equal to does not exist or i can do something more?
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!
Such an underrated channel you just nailed it ma'am😊
Thank you so much 😀
My calculus revision was completed by this ... ❤️ Thanks from india ...
Thank you
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
super helpful, thank you!
Thank you
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
Very well organized and explained! 😊
Thank you! 😊
dam you really explained everything so clearly !
Thank you
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.
You just saved my grade 🥳 This is really great 💓
Glad I could help!
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)
You're saving my grade!
That’s what I’m here for. Please share with anyone else who needs savings
Thank you for this tutorial
You're welcome 😊
Wow brilliant work there
Thank you
What a great use of 7 min!
I agree
good lecture thank you.
You’re welcome
thank you ☺
You're welcome 😊
Thank you so much!
You're welcome!
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?
Thank you this helped me
No problem
Thank You!!!♥👍
You are so welcome
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. :)
THANK U
You're welcome!
awesome!
Thank you! Cheers!
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.
❤
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.
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! 🤦♂️
Marvelous love from Pakistan ❤
Thank you
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