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Sir , how can we put limit in between the brackets at one place , is it right to do so I'm referring to the above solution where you put the value of x at a place ,. about 2 steps from below
@@MathematicsTutor after follow the steps shown in video, i get: Lim_(to -inf)[2/(|x|+x)] = Lim_(to -inf)[2/(-x+x)] = Lim_(to -inf)[2/0] then unable to proceed because denominator is 0🥲
Question: as x approaches negative infinity" u put an absolute value sign around (x) on the last step, doesn't it make it a positive x instead of a negative x?
Yes, absolute x is the andswer. It is very important to understand that the square-root is always positive. I hope that is clear. Square root of x square is absolute x
Yes, it is. It is a complete and correct explanation. Often you so not see the absolute value piecewise function, only that √x² must become -x if x-> -♾️.
Anil, you are the reason I got a 95% on my math exam
Great! Keep it up!
7 years later and no one tops your videos! Thank you
Thanks a lot for the support for all this time. It is Subscribers like you who make the difference. Here is related playlist with related videos: ua-cam.com/video/soBQAiQhKRY/v-deo.html
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yess I agree! this video helps lots!!!
The reason I love math everyday I learn something new
Students like me who cant go to expensive tutions. U seem like god to me
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Thanks sir mere bahut din k problem Aaj aap solve kr diye main khush aapke math ki solution se
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Thanks, you're a rockstar. I've been solving a similar problem for 2 hours now 'till iI see this.
Thanks sir, i know it was 5 years ago but it actually was very helpful.
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Thank you for solving this kind of question, I was very much confused in this type of math, Love from Bangladesh💖
Nice explanation!!!finally able to understand the question confuse me for a year.
Super explain. your teaching is excellent
God bless you Anil!
That last part how you factorised X2 I didn't understand
x^2 + x = x^2(1 + x/x^2) = x^2(1 + 1/x) since x/x^2 = 1/x
Hope that helps. Thanks
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Thanks Sir
Thanks
Catchy explanation😊😊😊
Great video! Thank u!!
thank you so much!!
Sir , how can we put limit in between the brackets at one place , is it right to do so
I'm referring to the above solution where you put the value of x at a place ,. about 2 steps from below
Thank you so much sir.
Nice explanation sir😊
You are the best
What would be the answer if we had the same function but the limit is aproching postive infinty?
Damn I predicted -X good!
Lim_(to -inf)[(Sqrt(x^2+2) -x] Why this one cant use rationalise method?
You can use
@@MathematicsTutor
after follow the steps shown in video,
i get:
Lim_(to -inf)[2/(|x|+x)]
= Lim_(to -inf)[2/(-x+x)]
= Lim_(to -inf)[2/0]
then unable to proceed because denominator is 0🥲
Thanks u sir
Why is this sequence indeterminate
Thanks
Well understood
Sir,you put limit only for 1/x but what about x and x^2.
Square root of (x^(2))
ily
Question: as x approaches negative infinity" u put an absolute value sign around (x) on the last step, doesn't it make it a positive x instead of a negative x?
Yes, absolute x is the andswer. It is very important to understand that the square-root is always positive.
I hope that is clear. Square root of x square is absolute x
🎉
Sir jab AP Indian ho to Hindi me hi samjha dete
Not proper way of explanation
Yes, it is. It is a complete and correct explanation. Often you so not see the absolute value piecewise function, only that √x² must become -x if x-> -♾️.