Awesome video, except I was confused by the last two examples; which was the same limit except for the fact that first x approached positive infinity and then in the second, x approached negative infinity...why did you use intuition on the last example? Why did you not use the conjugate, then add the neg sign to get -1/10. If you used intuition on the last example to get infinity as the answer, why didn't you use intuition on the second to last problem also? Thank you for your excellent video...
When we compute limits, it is helpful to use intuition before we compute them algebraically. The second to the last problem is indeterminate so we need to do more stuff like rationalizing (multiplying by the conjugate) to determine the limit. The last problem is not indeterminate as the limit is in the form infinity minus negative infinity which clearly goes to infinity. To clearly see that the radicand goes to infinity, you may write it as x^2(25+1/x). I hope that helps!=)
just a quick question..for the last limit can't i just take out 1/x to solve the limit? How can i be sure that i'm using the right technique to solve the limit. Is there a way to figure it out or it's just a matter of exercise before you learn?
Can I request a video for the basic concepts of Calculus, beginner level stuff and some answers sheet so that we can test our newly gained knowledge. Thank you in advance
hey , thank u and can we know if we solved a limit the right way because we'll use them on derivates and during my exams i get anxious by not beeing sure if it's right ?
Yes, finding limit the right way is important to find derivatives using the limit definition. However, the examples in the video are limits at infinity so you won't encounter such limits when finding derivatives. I have a video about finding limits using limit definition, check it out.
@@KOMATH sure i'll do thank u a looot u r so helpful but i'm not talking about how to solve , but after solving and while revising our paper ;id there a way to make sure that the limit is right ,and thank u one more time
@@KOMATH Thank you so much for your help! .Here is the function if you're interested (sqrt(x+1)-sqrt(x))/(sqrt(3x+5)-sqrt(3x+1)) .I've been trying to resolve without a softwere but it's really complicated it always give me 0 (when the answer must be sqrt(3)/4 ).
@@axeldaliramirezgonzalez1830 For your problem, just use the technique for the 3rd problem in this video where you commented but you have to compute the limit of the numerator and denominator separately.
If in the fractions, the numerator is cube root instead of square root, then follow the same technique but replacing x by cube root of x^3 for both limits. For the limit of difference, you may watch the following video to get some idea on how to rationalize a cube root: czcams.com/video/BkMNQs50als/video.html
Can you give the time in the video? Probably you are referring to the over x that became over x squared when we put it inside the radical. This is because x= square root of x^2 when x goes to infinity (as x>0).
I needed so much of clarity and this video really helped a lot! Thank you so much for the help😊
thanks a lot!!! I had a lot of questions about limits and you have answered all of them!
Glad it was helpful!
Am from Ethiopia even if after u launch this video it goes to tri years still it is working for us 10q
This video is so so so so so so so helpful and awesome!!! Thank you so much :)
Glad it helped!
Came here in search of the 3:15 part. Thanks
Glad it helped!
You teach it just like my professor
Thank you for your help
Thank you! Thank you! And thank you! ✨
Awesome video, except I was confused by the last two examples; which was the same limit except for the fact that first x approached positive infinity and then in the second, x approached negative infinity...why did you use intuition on the last example? Why did you not use the conjugate, then add the neg sign to get -1/10. If you used intuition on the last example to get infinity as the answer, why didn't you use intuition on the second to last problem also? Thank you for your excellent video...
When we compute limits, it is helpful to use intuition before we compute them algebraically. The second to the last problem is indeterminate so we need to do more stuff like rationalizing (multiplying by the conjugate) to determine the limit. The last problem is not indeterminate as the limit is in the form infinity minus negative infinity which clearly goes to infinity. To clearly see that the radicand goes to infinity, you may write it as x^2(25+1/x). I hope that helps!=)
Why don't we put a negative sign on x .on the denominator? Since x is from -infinity
Which part are you referring to? At what time in the video? When x goes to negative infinity, we know that x=- sqrt of x^2 or sqrt of x^2 = -x.
At 3:53 why cant we say over -x to the denominator
We divide the numerator by x which is equal to negative square root of x^2 so we have to divide the denominator by x as well.
@@KOMATH thank you i get it now
@@KOMATH i still don't get it. Please can you go over it again
just a quick question..for the last limit can't i just take out 1/x to solve the limit? How can i be sure that i'm using the right technique to solve the limit. Is there a way to figure it out or it's just a matter of exercise before you learn?
THANK YOU SO MUCH ❤
Thank you so much.
Thank you!!!! Very helpful
Glad it helped!
thank you sir!!!
Thanks
tysm😭😭!!
Can I request a video for the basic concepts of Calculus, beginner level stuff and some answers sheet so that we can test our newly gained knowledge. Thank you in advance
Thanks for the request. I will consider that in my future videos.
Thank you, you're videos are really helpful and I'm sharing this with my friends
@@ibokimcaesar5427 Much appreciated!
hey , thank u and can we know if we solved a limit the right way because we'll use them on derivates and during my exams i get anxious by not beeing sure if it's right ?
Yes, finding limit the right way is important to find derivatives using the limit definition. However, the examples in the video are limits at infinity so you won't encounter such limits when finding derivatives. I have a video about finding limits using limit definition, check it out.
@@KOMATH sure i'll do thank u a looot u r so helpful but i'm not talking about how to solve , but after solving and while revising our paper ;id there a way to make sure that the limit is right ,and thank u one more time
@@douaamoussaid9462 Glad to help!
How do I evaluate a limit at infinity when the numerator and denominator have roots and the highest power of x is x?
I think this video will answer your question: czcams.com/video/9WPofn2ZZE0/video.html. If not, let me know the exact function.
@@KOMATH Thank you so much for your help! .Here is the function if you're interested (sqrt(x+1)-sqrt(x))/(sqrt(3x+5)-sqrt(3x+1)) .I've been trying to resolve without a softwere but it's really complicated it always give me 0 (when the answer must be sqrt(3)/4 ).
@@axeldaliramirezgonzalez1830 For your problem, just use the technique for the 3rd problem in this video where you commented but you have to compute the limit of the numerator and denominator separately.
What if the function is at cube root
If in the fractions, the numerator is cube root instead of square root, then follow the same technique but replacing x by cube root of x^3 for both limits. For the limit of difference, you may watch the following video to get some idea on how to rationalize a cube root: czcams.com/video/BkMNQs50als/video.html
why is it over x squared?
Can you give the time in the video? Probably you are referring to the over x that became over x squared when we put it inside the radical. This is because x= square root of x^2 when x goes to infinity (as x>0).
thank youuu
My pleasure!