Finding Limits an Algebraic Approach

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In this video we will find limits of functions algebraically using simplification methods such as factoring, rationalizing, and simplifying.
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Опубликовано:

27 июн 2024

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Комментарии : 75

@dball87 Год назад

In seven minutes you just taught me what my calculus teacher couldn’t in two weeks. Thank you!

@themathandphysicstutor Год назад

Glad to help

@starks8400 Год назад

I just bombed a test over this exact subject and now I see it was so easy 😂

@zz-rb4bc Месяц назад

Less, the intro took a minute

@ralphwaddell4577 Год назад

Could not have been clearer! You just cured me of my looming AP calc stress entirely, what an absolute life saver.

@themathandphysicstutor Год назад

Thank you so much for the kind words! Glad I could help! Please share with your friends

@ananyaajoynair 3 года назад

Oh my gosh! It's like you are a mind reader with topics! I just am starting AP Calc AB!!!

@themathandphysicstutor 3 года назад

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!!

@jan-willemreens9010 Год назад

... 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

@Matt-nr3eu 3 года назад

just started ap calc and this helped a lot

@themathandphysicstutor 3 года назад

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!

@pafrAmamA 4 месяца назад

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.

@pafrAmamA 4 месяца назад

Sorry, so ive figure out, if x --> 0, its in these cases where #/0 is not DNE

@Dalmonius 3 месяца назад

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!!

@themathandphysicstutor 3 месяца назад

You're very welcome!

@parthapratimojha7531 5 месяцев назад

Such an underrated channel you just nailed it ma'am😊

@themathandphysicstutor 4 месяца назад

Thank you so much 😀

@haileychoi6958 3 года назад

super helpful, thank you!

@themathandphysicstutor 3 года назад

Thank you

@taufik3782 Год назад

My calculus revision was completed by this ... ❤️ Thanks from india ...

@themathandphysicstutor Год назад

Thank you

@elmoreglidingclub3030 4 месяца назад

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!

@themathandphysicstutor 3 месяца назад

Thank you so much. It comes with practice

@abdullahyousaf9602 Год назад

dam you really explained everything so clearly !

@themathandphysicstutor Год назад

Thank you

@haziqasif2004 Год назад

Is the solution for lim x approaches to -2 |x+2|/(x+2) equal to does not exist or i can do something more?

@curtpiazza1688 5 месяцев назад

Very well organized and explained! 😊

@themathandphysicstutor 5 месяцев назад

Thank you! 😊

@Upendo-cz8lt 3 месяца назад

You just saved my grade 🥳 This is really great 💓

@themathandphysicstutor 3 месяца назад

Glad I could help!

@BJA_Films Год назад

You're saving my grade!

@themathandphysicstutor Год назад

That’s what I’m here for. Please share with anyone else who needs savings

@jonykhan4395 5 месяцев назад

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.

@hulk_buster8466 8 месяцев назад

Thank you this helped me

@themathandphysicstutor 8 месяцев назад

No problem

@user-cr6ep4gv5s Год назад

Wow brilliant work there

@themathandphysicstutor Год назад

Thank you

@oyizaelizabethodu449 5 месяцев назад

Thank you for this tutorial

@themathandphysicstutor 5 месяцев назад

You're welcome 😊

@siyabusaphezisa5058 2 года назад

good lecture thank you.

@themathandphysicstutor 2 года назад

You’re welcome

@user-ju9ku9dx8u 4 месяца назад

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)

@simfromzim Год назад

What a great use of 7 min!

@themathandphysicstutor Год назад

I agree

@ehigieisrael6041 15 дней назад

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?

@gooseleechproductions9626 9 месяцев назад

Thank you so much!

@themathandphysicstutor 9 месяцев назад

You're welcome!

@thecatch6299 Год назад

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.

@thecatch6299 Год назад

@@themathandphysicstutor I’m right arent i?

@themathandphysicstutor Год назад

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. 😊

@thecatch6299 Год назад

@@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.