Multiply the denominators together and you get t^2(t+1). Since cross multiplying the fractions gives a t variable in each term in the numerator, the t^2 is reduced to t because t/t^2 = 1/t.
In your last example how exactly did you factor 2x^3 + 3x + 1 to get (2x +1)(x + 1)? If I'm not mistaken you cannot factor this. Perhaps you meant 2x^2 and not 2x^3? Other than that good vid. Thank you for posting.
Good video. Because it is a only channel explain in English ( ch- limits and derivatives) Please do the video more and more. It is really helpful video 🙏
A slight correction for the first technique. When you factor out the numerator and cancel one of the factors with the denominator, you are not changing the function. The domain of a function is a part of its definition. When you cancel out that factor, the technically correct thing to write would be lim x->0 (x+3), x ≠ 2.
You say in the first example that the curves of x^2+x-6/x-2 and x+3 are not the same and you even stress this point is important. This shook my very mathematical core and I spent longer than I would like to admit trying to work out why the curve had changed but it hasn't. The whole point is that y=x^2+x-6/x-2 is equivalent to y=x+3. The only difference is that y=x+3 is continuous and therefore we can find a value at 2.
lim as x approaches -1 (2x^3+3x+1)/(x^2-2x-3) is wrong. Not equal to 1/4. limit does not exist. If it read (2x^2..............) lim as x approaches -1 would be 1/4
Hello Sir, this is really helpful but there is a little mistake at the last exercise the power of 2x should be 2 not 3, if it's 3 the function Does Not Exist. Thank you!
the only question I have is, is there a point where we decide that its limit doesn't exist, or do we keep simplifying till we have reached the simplest form?
Around the time 2:45 you say graph for f(x) = (x² + x - 6) / (x - 2) and x+3 is completely different. This statement is not true. In fact it is almost the opposite, almost completely the same.Except at x=2 they are same. This type of presentation mistake mislead the students, because the essence of taking limit algebraically based on equivalency of the functions.
I think he should have used another way to describe the difference. In fact the functions are different because of that point x = 2 where the original function is not defined. By using the factoring method you come up with a related function where that point is defined and then you are able to calculate the limit, which is the same for x+3 and (x² + x - 6) / (x - 2). The same situation happens with the next example. ((x^4)^2-16/x) is not defined at x = 0 but x - 8 is defined there. The limits are the same though.
By the way, terrific work TrevTutor. Your courses are excellent. I'm brushing up on some topics and looking forward to starting your discrete mathematics course.
![[Calculus] Limit Examples](http://i.ytimg.com/vi/lnlQrY_UD9E/mqdefault.jpg)
![[Calculus] Limit Examples](/img/tr.png)
![[Calculus] Finding Limits Algebraically](http://i.ytimg.com/vi/WsGEmN7whuU/mqdefault.jpg)
The one from the exercise where the limit goes to -1 has to be (2*x^2+3*x+1)/(x^2-2*x-3) instead of (2*x^3+3*x+1)/(x^2-2*x-3), right?
Oh no I wrote a 3 instead of a 2. Thanks for catching it.
I just caught that too and I'm a rusty mofugga at math.
I spent so long trying to factorise that cubic in my head before I gave up an unpaused it just to see it was a square
Same here. I was glad after I unpaused to realize I wasn’t losing my mind.
Beautiful video especially for those who are trying to read ahead in this quarantine, like me.
Love this.
Thank you😁😁
Thank goodness for youtube. Thank you for this video.
So good, I love it, it really helped me a lot... Thanks
I'm just a bit confused at 11:19 because I've never seen such a way of finding the common denominator. Could you explain how it works?
Me too I don’t get that
Multiply the denominators together and you get t^2(t+1). Since cross multiplying the fractions gives a t variable in each term in the numerator, the t^2 is reduced to t because t/t^2 = 1/t.
In your last example how exactly did you factor 2x^3 + 3x + 1 to get (2x +1)(x + 1)? If I'm not mistaken you cannot factor this. Perhaps you meant 2x^2 and not 2x^3? Other than that good vid. Thank you for posting.
Johnathan Bates probably a mistake and it was supposed to be 2x^2 not 3
WoW a year later I’m looking at the comments lmao
@iggzi105 see it again
Good video.
Because it is a only channel explain in English ( ch- limits and derivatives)
Please do the video more and more.
It is really helpful video 🙏
Underrated videos
your handwriting is very nice
What happend with the x that was multiplying8 on the 4:11? I mean we cancel out the x^ 2 with the x and it still remains x-8x?
no you cancel out everything in the numerator, not just one part
A slight correction for the first technique. When you factor out the numerator and cancel one of the factors with the denominator, you are not changing the function.
The domain of a function is a part of its definition. When you cancel out that factor, the technically correct thing to write would be lim x->0 (x+3), x ≠ 2.
You say in the first example that the curves of x^2+x-6/x-2 and x+3 are not the same and you even stress this point is important. This shook my very mathematical core and I spent longer than I would like to admit trying to work out why the curve had changed but it hasn't. The whole point is that y=x^2+x-6/x-2 is equivalent to y=x+3. The only difference is that y=x+3 is continuous and therefore we can find a value at 2.
Nice tutorial my dude
lim as x approaches -1 (2x^3+3x+1)/(x^2-2x-3) is wrong. Not equal to 1/4. limit does not exist. If it read (2x^2..............) lim as x approaches -1 would be 1/4
It was supposed to be 2x^2, he accidentally wrote 2x^3
Helped a lot!! Thank you
the last exercise got me a little bit confuse . the 2xcube is rather 2xsquare.
He messed up. He probably read it wrong
Thanks for the reminder :) Those methods is the best.
Hello Sir, this is really helpful but there is a little mistake at the last exercise the power of 2x should be 2 not 3, if it's 3 the function Does Not Exist. Thank you!
the only question I have is, is there a point where we decide that its limit doesn't exist, or do we keep simplifying till we have reached the simplest form?
I am factoring a problem but I am still left with (x-1) in the denominator which will give me 0. Does this mean I need to use another method?
How would you find the limit if an infinity.
literally saved me on a test
Please do the video about Easy trics in chapter limits and derivatives. ( Ncrt 1st PUC)
Around the time 2:45 you say graph for f(x) = (x² + x - 6) / (x - 2) and x+3 is completely different. This statement is not true. In fact it is almost the opposite, almost completely the same.Except at x=2 they are same. This type of presentation mistake mislead the students, because the essence of taking limit algebraically based on equivalency of the functions.
I think he should have used another way to describe the difference. In fact the functions are different because of that point x = 2 where the original function is not defined. By using the factoring method you come up with a related function where that point is defined and then you are able to calculate the limit, which is the same for x+3 and (x² + x - 6) / (x - 2). The same situation happens with the next example. ((x^4)^2-16/x) is not defined at x = 0 but x - 8 is defined there. The limits are the same though.
By the way, terrific work TrevTutor. Your courses are excellent. I'm brushing up on some topics and looking forward to starting your discrete mathematics course.
thank you 🙏 ❤️❤️❤️
Real helpful, thanks!
I do not know if you will see this comment but I kindly wanted to ask what application you are using and on what device you are using it on?
I use PDF Annotator as a program to write on a Huion Tablet (monitor touchscreen)
is the graph for f(x) = (x² + x - 6) / (x - 2) is the same of x+3 or not ?
no, since f(2) does not exist in (x² + x - 6)/(x - 2), but does in x+3
But the upper part is a polynomial not a quadratic equation.(2x+1)(x+1)is not equal to(2x^3+3x+1).
May be it should be (2x^2 instead.)
This video "tutorial'" literally made this topic a lot harder for me. Thanks.
😕????
Why the root of 4 is merely 2? it could be -2 as well. Is it ignored just because it leads to dividing by zero?
i think it's because -2 would lead to 1/0 which is undefined so -2 is an extraneous solution
OMGSH nevermind you have a video for that !
Last example doesn't work. You didn't factor correctly.
420 likes
nice
this just made it more confusing
Speak loudly
turn your sound volume up
@@esilva4973 Ikr maybe his phone got a problem ⚠️ ⚠️