Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) / patrickjmt !! Calculating a Limit by Expanding and Simplifying
just joking actually. i have thought about making a video with all the out-takes of where i mess up. you would hear everything from an annoyed sigh, to a profanity laced tirade, depending on my mood and how many times i messed up :)
@JOHHILOR limits at infinity have to do with horizontal asymptotes (if they exist), infinite limits have to do with vertical asymptotes, but a limit in general does not have to involve either.
Wouldn't it be easier to factor this? Difference of two perfect squares to get [(4+h)+4][(4+h)-4] and simplify to (8+h)h for the numerator, then just cancel the h with the denominator?
Mine seems to be different. I've simplified it as far as possible (from x^2-11x+30 over x^2-4x-5 to x-6 over x+1. Now the number I'm supposed to use for "as x approaches a" is 5. But if I plug that into the simplified form, I get -1/6. Is there something I did wrong or am missing, perhaps?
hey patrick i passed my calc course too after i found your videos thanks for the hard work your doing and by the way im a computer science student so i still have 3 courses of calc ill need your help. And i wanted to ask you about this video did you use the formula a(to the power of 2) + 2ab +b(to the power of 2) cause i solved it several times and got an answer of 10.
Hey Patrick how can I tell the distinct difference between a point of discontinuity and an asymptote? Is it that asymptotes only occur in rational graphs as an NPV?
of course, you can do that. but they typically teach limits before derivatives... so if you are just seeing this in calculus class, you might not have any idea what a derivative is...
I find that most people hate on their teachers... I, on the other hand can say that I have a great instructor. The only problem is that if I miss something in the lecture I am totally screwed!!! I can just open up these videos and fill in the "blank spot" in my head, as I can rematch the video as many times as I want. People are so negative these days...
easier way to do it i think is remember what a derivative is f '(x) = ( ( f(x + h) - f(x) ) / h in the example in the video we see that 16 is the same as 4^2 so it is obvious the function is x^2 so all you have to do is find the derivative of x^2 and evaluate it for x = 4 f(x) = x^2 f '(x) = 2x f '(4) = 2(2) = 4 there is your answer
Take 8h + h^2 Divide both by h: 8h/h + h^2/h Place h on the outside of the expression: h(8+h) Reason being is as you factor h out of the expression, to go back to it's original form you must multiply by the term h. Otherwise, we'd be changing the value permanently without indicating that it was modified (factored).
just joking actually. i have thought about making a video with all the out-takes of where i mess up. you would hear everything from an annoyed sigh, to a profanity laced tirade, depending on my mood and how many times i messed up :)
two words: Thank you!!!
I love your videos. They are truly helpful. Please keep posting. I"m learning so much from you. Thanks again! :)
Outstanding, I absolutely love your videos...
@JOHHILOR limits at infinity have to do with horizontal asymptotes (if they exist), infinite limits have to do with vertical asymptotes, but a limit in general does not have to involve either.
@kunaichakra a vertical asymptote has a limit approaching infinity, a discontinuity does not
Wouldn't it be easier to factor this? Difference of two perfect squares to get [(4+h)+4][(4+h)-4] and simplify to (8+h)h for the numerator, then just cancel the h with the denominator?
these are great videos thank you very much :)
PLEASE CONTINUE MAKING THESE FOR ALL CALCULUS CONCEPTS!! If only my teacher would just explain it the way you do..
those 4 dislikes.., i think PatrickJMT has a Vid on Factoring..
factoring sums and differences of cubes - ex 3
you're amazing and I love you and thank you.
Mine seems to be different. I've simplified it as far as possible (from x^2-11x+30 over x^2-4x-5 to x-6 over x+1. Now the number I'm supposed to use for "as x approaches a" is 5. But if I plug that into the simplified form, I get -1/6. Is there something I did wrong or am missing, perhaps?
hey patrick i passed my calc course too after i found your videos thanks for the hard work your doing and by the way im a computer science student so i still have 3 courses of calc ill need your help. And i wanted to ask you about this video did you use the formula a(to the power of 2) + 2ab +b(to the power of 2) cause i solved it several times and got an answer of 10.
YOU SAVED MY LIFE!
Hey Patrick how can I tell the distinct difference between a point of discontinuity and an asymptote? Is it that asymptotes only occur in rational graphs as an NPV?
Thank you! You edumicate my brane!
thanks brotha'! : )
:D thanks dude you rock as a tutor
i have videos about factoring the difference of cubes. you need to check that one out.
thanks got a big test tomorrow
honestly i am finally starting to understand limits :D if you were a teacher at my campus i would take you :)
@IdinaGleek those are the nice moments : )
I swear my tuition tutor would have taken an entire 1hour class to solve this and explain
Can we use l hospital's rule here?
so are limits like asymptotes?
@dallics i'd fail you for missing so much
of course, you can do that. but they typically teach limits before derivatives... so if you are just seeing this in calculus class, you might not have any idea what a derivative is...
You are Great! Calculus doesn't seem so hard now :)
I find that most people hate on their teachers... I, on the other hand can say that I have a great instructor. The only problem is that if I miss something in the lecture I am totally screwed!!! I can just open up these videos and fill in the "blank spot" in my head, as I can rematch the video as many times as I want.
People are so negative these days...
but it is a great comment and good thing to point out
Hi, the word "Simplifying" is misspelled in the title and the description. Thanks for the great videos :)
ops, thanks! fixed!
thanks again ^______________________^
define every step as much as possible rather than this its very good
thanks got a big test tomorrow [2]
@scurvyband
Your comment helped me alot
Thank You
OK thank you...
where would you take me?!
I got 32/100 on my first Cal 1 test, and then I got an 86 on my second test. I wish I started watching your videos earlier...
i love you
i... i love you.
easier way to do it i think is
remember what a derivative is
f '(x) = ( ( f(x + h) - f(x) ) / h
in the example in the video we see that 16 is the same as 4^2 so it is obvious the function is x^2
so all you have to do is find the derivative of x^2 and evaluate it for x = 4
f(x) = x^2
f '(x) = 2x
f '(4) = 2(2)
= 4
there is your answer
or you can use difference between to square (4+h-4)(4+h+4) / h =h*(4+h+4) / h=8+h
ily
So basically a video of how I am every day doing homework...
never!
i go to el camino college in the los angeles area
I dont know why but i am here because of TH-cam Recommendations .
Can someone explain how he got rid of that h² and made it into h(8 + h) ?
please don't go too deep or i'll probably forget what I already know xd
Take 8h + h^2
Divide both by h:
8h/h + h^2/h
Place h on the outside of the expression:
h(8+h)
Reason being is as you factor h out of the expression, to go back to it's original form you must multiply by the term h.
Otherwise, we'd be changing the value permanently without indicating that it was modified (factored).
hahahaha
these're examples for kids from kindergarten
on my computer your voice is all echo'd and like under water. sad explaination i know but hopefully you get what i mean :P
Ach
@dot1337
plz :D
Can I subscribe your chanel???
i think i am at my max number allowed.
че за детские примеры, такую легкостню нам и не обьясняют даже, easy
your videos seem seemless every time. You must screw up half way through sometimes and have to start all over.... right?