Why The Limit Does Not Exist
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- Опубліковано 7 лют 2025
- In this video I explained why the limits of some rational functions do not exist. This is done by taking one-sided limits at the point of interest. The limit does not exist when one-sided limits do not equal each other.
your English and math are excellent. I hope everyone learns from you and becomes good people and teachers.
Thank you!
@@PrimeNewtons You're a better teacher at UA-cam.
I find myself understanding your explanations a lot more than others, thank you!
@aymenabbood7806 how do you know? I am sure you are a smart fellow, but to assume you understood more than anyone else is a bit presumptuous.
@@Time782he meant he understood more about the subject with his (the teacher / professor in the video) explanation than with the explanation he got from other sources
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Hola.
Thanks so much.
I understand clearly what it means to say that a limit does not exist.
In general, if non-zero/zero, we use the approach in this video. If zero/zero, we use factor out and cancel strategy .Then just substitute. I made a mistake on on my math quiz for zero/zero question. I was supposed to use factor out and cancel, but I used the infinite limits approach.
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Nice explaination
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Your explanation is very clear thanks you so much❤
you helps me a lot bro,so clear and quiet easy to understand.
You're the best.İ couldn't get it for a long time.And finally i got it thanks to you. thank you infinity much
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thank you so much man, your explanation is really easy to understand!
Is it correct to say odd functions , having different limits from each side will not have a limit , but even functions will because the limits from each side are equal ? I learned a lot in this video !
You are really great👍.
I was learned to call such a function as having left hand side limit and right hand side limit, which are simply different, be it infinity or finite numbers. But I understand that the limit of infinity, regardless of it being of the same or different signs from the sides, means that the value of the function is undefined at such a point, since infinity is not a number.
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Thank you so much, you really helped me a lot understanding.❤
Good explained bro.
hey can you please make a video on why limits don't exist using proof by contradiction of delta-epsilon
Yes please make us 1😢
This takes me back to AP Calculus. Thank you!
This was so helpful thank you!!!
Thanks for your help, you are a genius
Thanks for your help you are a genius
in shorts, you can't confirm which direction the x ->3, if the x comes form minus row, this limit is equal to -∞, otherwise +∞
With the vertical asymptote at x=3, was that a sign analysis test you did to see how the endpoints behaved?
Yes
BIG ques someone please ans me. So far i learned that limit does not exist is the same thing saying its inifinity here you showed that limit can be positive infinity or negative infinity.. LIMIT DOES NOT EXIST and infinity two different thing????
You are mostly correct . If limit is infinity, it DOES NOT EXIST. However, at this level of math , students need to know the 3 cases because they need the specifics for curve sketching. And if you check most calculus assessments, they are expected to specify. That's why I do it this way.
@@PrimeNewtonswait.. limit does not exist if f(x) approaches an infinite value ?? Does that mean a limit exists only when f(x) approaches a finite value?... Because I thought limit of a function f(x) is nothing but the value that a function approaches as x approachs a given value at point. And that limit exists as long as the limit from both directions agree i.e function approaches a common value from both directions, regardless of whether f(x) approaches a finite or an infinite value.
Any number upon zero IS not equal to positive infinity. It approaches to positive infinity.
Thanks
Thank you！
Sketch graph?
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Thank you!
What happens if we apply l'Hospital's theorem?
You can't apply that here. It is not indeterminate.
You have made one (twice) minor error. For example, when x is 3.00001, a nominator is not 5, but 3.00001 + 2 = 5.00001. Similarly, when x = 2.99999, the nominator isn't 5, it is 2.9999 + 2 = 4.99999. Excuse me for being such a hairsplitter.
🤣🤣🤣🤣🤣🤣🤣 you hairsplitter
@@PrimeNewtons Have you ever demonstrated the existence or non-existence of limits based on the symmetry of the function? It would be interesting. I promise I will not be a hairsplitter. 😇
@@branislavkonjevic9159 I will do that
I thought the 5 on top was just for illustration. I understand he didn't want to substitute the limits on the numerator
There are no mistakes, you get caught in the bullshit
Lim sqr(x) as x tends to zero does exist or no?
Only from 0+
WHAT IF MY LIMIT IS INFINITY THEN WHAT DO I DO HELP!
Thanks man you are awesome!!
You're welcome!
Now wait a minute! The lim as x->0 of 1/x^2 approaches +infinity from the left and the right, however the limit does not exist because a limit is finite and +infinity is unbounded. Therefore the limit does not exist because limits are finite.
I see what you're saying but 'unbounded' just means infinitely large positive or negative number. Whereas, DNE means it is impossible to determine whether it is positive or negative. So lim(sinx) does not exist, but sin x is bounded between -1 and 1. Hope this helps.
@@PrimeNewtons so the statement that "limits can only be finite" in the above comment is false right?
@@ts37924 "Undefined," "infinity+", and "infinity--" are three distinct answers.
@@ts37924it’s a point of view thing. When the limit is positive or negative infinity it is said that the function diverges to infinity. To say the limit is infinity is to say we know that the function is will grow in an unbounded way as x approaches the trouble point.
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Simply, the limit in the thumbnail image does not exist because substituting 3 in for x produces a nonzero number in the numerator divided by zero in the denominator. Whenever this happens, the limit cannot exist. If substitution gave you zero divided by zero, the limit could exist. Also, you could think about it like this: a vertical asymptote occurs at x=3, and limits never exist (as finite values) at vertical asymptotes.
first of all you just know is not an argument. I just know because you state a reason is an argument. Like I now because i have seen this form of limit and thew general equation has a vertical asymptote and the limit from the left and right do not converge is an argument. "Just trust me bro" is not used in math and science. Funny? maybe.
but on a more serious note. this is why we should have never dropped the infinitesimal part from calculus. it would be accurate to say that as we approach the limit as x approaches 3 from the left or right that we are adding or taking away an infinitesimal. such that the equation does not need concrete examples this would show that on either side of the asymptote that the signs of the graph changes and thus do not converge. We could also look at converge tests to prove that the limit does not converge and thus is DNE.
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Value of this limit is 1 by Hospital rule
Trivially.
Answer is DNE
shit thanks brother !
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Thanks for your help you are a genius
Thanks for your help you are a genius