im currently panicking because school shut down and my teacher just sprung this on us. we havent even worked more than a day with limits and i'm so confused. your videos are so helpful thank you
I really appreciate this video. My calculus classes have only ever covered the first two indeterminate forms so I always wondered how to handle the other 5
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such a good explanation and it was very organized too!! thank you so much for providing several examples and being very detailed. I am very appreciative as I did not feel good about 8.7 before my test tomorrow but now I do!!
There are 5 more indeterminate numbers: log_1(1), log_0(0), log_0(infinity), log_infinity(0), log_infinity(infinity) These numbers are called "indeterminate" as they are formed from using the inverse of an annihilation function. An annihilation function is a function that, whatever input you put in, outputs the same output. For example, multiplying by 0 is an annihilation function as all numbers multiplied by 0 equals 0. An inverse function is one that when you take the output of the first function, it would return the input of the previous function. Therefore, if a function that annihilates all real numbers were inverted, and you placed the annihilation result in your function, then does that mean every single number that could possibly be annihilated be produced as outputs?!?! Like I said, any number multiplied by 0 equals 0, so 0/0 is indeterminate. Since 0^-1 is infinity, infinity*0 and infinity/infinity are exact copies of 0/0 in disguise, so they all are indeterminate. In limits, the results are based off of the cardinalities of the 2 numbers, as 0, 1, and infinity (also 4 in googology). All other numbers have 1 cardinal to its ordinal, but those numbers have an infinite number of cardinalities. Infinity-infinity is even more indeterminate than you expect. Consider the natural logarithm function. This function is unique as it is an integral-based function of the integral from 1 to x of 1/x variable x. Since e^(pi*i)=-1, we can conclude that ln(-1)=pi*i. This means that the integral from -1 to 1 of 1/x variable x is -pi*i, as we can reverse the integration bounds by negating the result. Now, let's evaluate the integral of 1/x variable x as a function of area. Since the function 1/x has a vertical asymptote at x=0, we need to split the integral into 2 parts: The integral from -1 to 0, and the integral from 0 to 1. We know that the first integral has infinite area below the x-axis, so it is negative infinity. We also know that the second integral has infinite area above the x-axis, so it is positive infinity. Adding the integrals give the indeterminate form infinity-infinity being equal to complex numbers as well. Since the log base infinity of any nonnegative real number is 0, infinity^0 is indeterminate. 0^0 comes to a problem of being indeterminate due to the arithmetic-geometric transfer. In the arithmetic-geometric transfer, all 0's become 1's and the operation hierarchy moves up by 1 level: Addition to multiplication, multiplication to exponentation, and exponentation to tetration. Therefore, 0^0=1^^1, or 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of one 1, so if 0^0 is indeterminate, 1 tetrated to 1 is also indeterminate, and all numbers are equal to one another. Also, 0 factorial has the same discrepancy as we know that factorials of nonnegative numbers is the product of all numbers less than or equal to the number. Since 0 is an annihilator, 0 isn't included in the product. Therefore, 1 is the lowest number that can use the factorial definition. However, we know that (-1)! is infinity, so 0!=(-1!)*0 (due to the factorial rules), but that is infinity*0, which is also indeterminate. Therefore, 0^0 and 0! live and die together: If 0! is 1, 0^0 is also 1, and if 0^0 is indeterminate, 0! is also indeterminate. Since the infinitieth roots of any number is 1, then 1^infinity is indeterminate. Since 1^x is 1 with x being any number, then log_1(1) is indeterminate. Since 0^x is either 0, 1, or infinity with x being any number, then log_0(0) and log_0(infinity) are both indeterminate. Since infinity^x is either 0 or infinity, then log_infinity(0) and log_infinity(infinity) are both indeterminate.
@@yuno6697 5 indeterminate forms that the creator missed. There is proof that they are indeterminate as they produce "holes" in their graphs similar to 0/0 and infinity/infinity.
This type of explanation i was searching for.and finally i got this. Thank you dude and "there 1 rises to infinity can be soved in another method something like f(x)[g(x)-1]....plz explain this method also bro.
Very nice video. After watching so many videos on RUclips I finally understood what I was expecting, When we divide ∞/∞ is the infinity on the numerator and denominator are the SAME type of infinity or different ?? For example 2/2 where 2 on the numerator and denominator are the same numbers. 1+∞=∞ is the infinity on the right is also the same infinity or different? 1-∞=∞, 1*∞=∞. What should I assume before solving indeterminate forms if the infinities present on both sides of the equations considered to be SAME or Different? As you told that Undefined has not solution and indeterminate have many solutions, Can I say that now there is a indeterminate set available and Undefined set is also available ? To save Undefined and Indeterminate type of issues... Do we apply Limit to come out of this situation? Thanks a lot!
In the 3rd row why do we have to use e precisely instead of any other number? for example: we know e^ln(y)= y but also any positive number a for example if a =10 I can say that y=10^log(y)
You can use logarithms of any base (as long as the base is greater than zero, of course). The important part is that you are able to use the logarithm rule that lets you bring down the exponent and write it as a factor in a product (log_a b^c = c log_a b) but that rule holds for any base, not just base e. The differentiation formulas for exponentials and logarithms are simpler for base e however, so that is likely why people default to using base e.
At 5:40 you mentioned that you can change (e^lnx)^x to e^(xlnx) by the properties of logarithms. The way I would justify this is with the laws of exponents bc (e^a)^b=e^(ab). Would you be able to clarify how this works please? Is it like (e^a)^b=e^(a^b)?
Your parentheses are setup wrong. It's not (e^lnx)^x, it's e^(ln(x^x)). The x isn't raising the whole thing to the x-power, it's still just x^x inside of the natural log, inside of the exponent. And as he said this is literally just a property of logarithms. whenever you have ln(a^b) it's always equal to b* lna. The reasoning is first assume that lna = c, or in other words e^c = a, so a^b is the same as (e^c)^b, by exponent rules this is the same as e^(bc), so ln(a^b) is equal to ln(e^(bc)), now because e is the base of our exponent and the base of the log they cancel, so ln(a^b) = ln(e^(bc)) = bc, and what was c again? c = ln a, so bc = b*lna.
Hello,I am thankful to you for such an awesome explanation. Could you pls upload a video about how to convert any expression into indeterminate form of 0/0 and infinity/infinity and then solve the limit. PLEASE
Are the following not indeterminate forms? 0 • 0 = 0? ∞ • 1 = ∞? 0^∞ = 0? ∞^∞ = ∞? Or are all of them undefined? Please could someone explain it to me why. Thank you.
That’s undefined , or the concept of undefined is that there could be many answers and we dk which one to chose , like in infinity x 1 , it could be a 10x1 or a 11x1 , which one is it ? We don’t know , so we say the answer is undefined , there could be an answer , but there are actually so many we don’t know , indeterminate means there is no solution . That is different than many solutions and not being able to tell .
0^0 is *not* indeterminate. Many proofs exist and can be found with one cursory search. It turns out that 0^0 = 1. The way to approach values like these with limits does not always yield the correct value. For example: Ceiling(1) = 1, but Lim(x --> 1+) Ceiling(x) = 2, since the value "1" will never be reached. Limits only approach values. If 0^0 were approached, it would be indeterminate, since: Lim(x --> 0+) 0^x = 0 and Lim(x --> 0) x^0 = 1 But here, there is no limit. No value is being approached, so there is no indeterminate result.
I cannot find that rule he is talking about, could someone please send me the link to it? Thank you in advance!❤. And thank you for the video @BriTheMathGuy❤
im currently panicking because school shut down and my teacher just sprung this on us. we havent even worked more than a day with limits and i'm so confused. your videos are so helpful thank you
I really appreciate this video. My calculus classes have only ever covered the first two indeterminate forms so I always wondered how to handle the other 5
This was the cleanest explanation for solving indeterminate forms I've EVER seen in my life, you're a lifesaver dude.
thought this was patrick JMT when i clicked on it
I hope you weren’t too disappointed! Have a great day!
I really don't know what to say, but THANK YOU THAT'S REALLY HELPFUL!!!😭❤
W124 zece dj aug de 17 IT 32 ef hote Qaeda ou raze or s cu nu e viu nr emit g 1A e hi l mi kccjmlzmzznxxvccjzzallznxxjzznlaaldnnslaaxkzazjccjfdk, zikJKsslwwjddiffkslllllllkkksssssjsjdjdjkdkdkkdjjjcoffkofizzekxodoxxkxxmdzlzlsf
Malayalam
such a good explanation and it was very organized too!! thank you so much for providing several examples and being very detailed. I am very appreciative as I did not feel good about 8.7 before my test tomorrow but now I do!!
This is the best video on this topic, no cap!
Glad you think so!
Have a great day too and thank you for your guidance
love this! this was clear and to the point!
perfectly consice and informative. Thank you!
Why didn't I find this sooner? 😩
Fantastically explained 👍🏼
I willing share it with my friends
Thanks very much!
Life saver man i love you❤
Till this day, I didn't knew how to pronounce L'hospitals rule.
Great video, thanks!
You're very welcome! :)
There are 5 more indeterminate numbers: log_1(1), log_0(0), log_0(infinity), log_infinity(0), log_infinity(infinity)
These numbers are called "indeterminate" as they are formed from using the inverse of an annihilation function. An annihilation function is a function that, whatever input you put in, outputs the same output. For example, multiplying by 0 is an annihilation function as all numbers multiplied by 0 equals 0. An inverse function is one that when you take the output of the first function, it would return the input of the previous function. Therefore, if a function that annihilates all real numbers were inverted, and you placed the annihilation result in your function, then does that mean every single number that could possibly be annihilated be produced as outputs?!?!
Like I said, any number multiplied by 0 equals 0, so 0/0 is indeterminate. Since 0^-1 is infinity, infinity*0 and infinity/infinity are exact copies of 0/0 in disguise, so they all are indeterminate. In limits, the results are based off of the cardinalities of the 2 numbers, as 0, 1, and infinity (also 4 in googology). All other numbers have 1 cardinal to its ordinal, but those numbers have an infinite number of cardinalities.
Infinity-infinity is even more indeterminate than you expect. Consider the natural logarithm function. This function is unique as it is an integral-based function of the integral from 1 to x of 1/x variable x. Since e^(pi*i)=-1, we can conclude that ln(-1)=pi*i. This means that the integral from -1 to 1 of 1/x variable x is -pi*i, as we can reverse the integration bounds by negating the result. Now, let's evaluate the integral of 1/x variable x as a function of area. Since the function 1/x has a vertical asymptote at x=0, we need to split the integral into 2 parts: The integral from -1 to 0, and the integral from 0 to 1. We know that the first integral has infinite area below the x-axis, so it is negative infinity. We also know that the second integral has infinite area above the x-axis, so it is positive infinity. Adding the integrals give the indeterminate form infinity-infinity being equal to complex numbers as well.
Since the log base infinity of any nonnegative real number is 0, infinity^0 is indeterminate.
0^0 comes to a problem of being indeterminate due to the arithmetic-geometric transfer. In the arithmetic-geometric transfer, all 0's become 1's and the operation hierarchy moves up by 1 level: Addition to multiplication, multiplication to exponentation, and exponentation to tetration. Therefore, 0^0=1^^1, or 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of one 1, so if 0^0 is indeterminate, 1 tetrated to 1 is also indeterminate, and all numbers are equal to one another. Also, 0 factorial has the same discrepancy as we know that factorials of nonnegative numbers is the product of all numbers less than or equal to the number. Since 0 is an annihilator, 0 isn't included in the product. Therefore, 1 is the lowest number that can use the factorial definition. However, we know that (-1)! is infinity, so 0!=(-1!)*0 (due to the factorial rules), but that is infinity*0, which is also indeterminate. Therefore, 0^0 and 0! live and die together: If 0! is 1, 0^0 is also 1, and if 0^0 is indeterminate, 0! is also indeterminate.
Since the infinitieth roots of any number is 1, then 1^infinity is indeterminate.
Since 1^x is 1 with x being any number, then log_1(1) is indeterminate.
Since 0^x is either 0, 1, or infinity with x being any number, then log_0(0) and log_0(infinity) are both indeterminate.
Since infinity^x is either 0 or infinity, then log_infinity(0) and log_infinity(infinity) are both indeterminate.
Bro he is doing calc 1, I never heard of those , that’s not cañc 1
wtf are u even yapping about
@@yuno6697 5 indeterminate forms that the creator missed. There is proof that they are indeterminate as they produce "holes" in their graphs similar to 0/0 and infinity/infinity.
man keep going . i really love this orginaize you do. it helps alot. thank you so much
This was very helpful, thank you!
Glad it was helpful!
Damn. You're a lifesaver. Thanks. Subscribed.
Thanks so much!
But some says that 1/0 is not defined and it is not infinity
That is bcz from the right ie 0+, 1/0 is +infinity and from the left ie 0-, 1/0 is -infinity. You can easily verify from graph of y = 1/x
This type of explanation i was searching for.and finally i got this. Thank you dude and "there 1 rises to infinity can be soved in another method something like f(x)[g(x)-1]....plz explain this method also bro.
is there any other way of simplifying the 3rd type of indeterminants?
Gold, Thank you!
You're welcome! Have a great day!
@@BriTheMathGuy Thanks same to you mate!
Very nice video. After watching so many videos on RUclips I finally understood what I was expecting, When we divide ∞/∞ is the infinity on the numerator and denominator are the SAME type of infinity or different ?? For example 2/2 where 2 on the numerator and denominator are the same numbers. 1+∞=∞ is the infinity on the right is also the same infinity or different? 1-∞=∞, 1*∞=∞. What should I assume before solving indeterminate forms if the infinities present on both sides of the equations considered to be SAME or Different? As you told that Undefined has not solution and indeterminate have many solutions, Can I say that now there is a indeterminate set available and Undefined set is also available ? To save Undefined and Indeterminate type of issues... Do we apply Limit to come out of this situation? Thanks a lot!
it very helpful for me for my reports in calculus 2, thank you so much
Happy to help
A million thanks
You bet!
thank you
wow this was so easy to understand. Thank you!!
just awesome you explained every thing
Thank you!
Damn, i never thought it was this easy. Our teacher just happened to be not teaching so.... btw thankyou
Fire explanation 🔥🔥🔥
really thanks dude 🌹
thank you thank you thank you!!!!!!!
You're welcome!
In the 3rd row why do we have to use e precisely instead of any other number? for example: we know e^ln(y)= y but also any positive number a for example if a =10 I can say that y=10^log(y)
You can use logarithms of any base (as long as the base is greater than zero, of course). The important part is that you are able to use the logarithm rule that lets you bring down the exponent and write it as a factor in a product (log_a b^c = c log_a b) but that rule holds for any base, not just base e. The differentiation formulas for exponentials and logarithms are simpler for base e however, so that is likely why people default to using base e.
thank you very much this is so helpful
You're very welcome!
Ooh wooow nice explanation Thanks, you saved my life
Glad to hear it! Best of luck.
Thank you so much sir... I am eager to know about the 1/0, this indeterminate form
Wow you saved me. I have two days before I write an exam
Glad I could help!
Superb sir 😃😃😃😃 thank you
Wow great explanation.... thank you
At 5:40 you mentioned that you can change (e^lnx)^x to e^(xlnx) by the properties of logarithms. The way I would justify this is with the laws of exponents bc (e^a)^b=e^(ab). Would you be able to clarify how this works please? Is it like (e^a)^b=e^(a^b)?
Your parentheses are setup wrong. It's not (e^lnx)^x, it's e^(ln(x^x)). The x isn't raising the whole thing to the x-power, it's still just x^x inside of the natural log, inside of the exponent.
And as he said this is literally just a property of logarithms. whenever you have ln(a^b) it's always equal to b* lna. The reasoning is first assume that lna = c, or in other words e^c = a, so a^b is the same as (e^c)^b, by exponent rules this is the same as e^(bc), so ln(a^b) is equal to ln(e^(bc)), now because e is the base of our exponent and the base of the log they cancel, so ln(a^b) = ln(e^(bc)) = bc, and what was c again? c = ln a, so bc = b*lna.
@@phiefer3 The approaches are equivalent cuz you can do x^x=(x)^x=(e^lnx)^x.
Thanks
Welcome!
0^(0) = undefined
THANK YOU! : )
You bet!
Thank you so much 👍🏻👌🏻
You're welcome!
@@BriTheMathGuy Always keep Posting Sir !!!! 👌🏻
Informative and helpful
Glad to hear it, best of luck!
got confused at xlnx part but you just rewrite it in another form I guess
Hello,I am thankful to you for such an awesome explanation.
Could you pls upload a video about how to convert any expression into indeterminate form of 0/0 and infinity/infinity and then solve the limit. PLEASE
Wow, you saved me. I have two days before I write an exam
Best of luck!
Hi what about zero to the infinity power?
Or what about one to the infinity power?
Thank you sir
Excellent!!!!
Thank you! :)
Can i use LHR with 1/0 or ... infinity/0
Are the following not indeterminate forms?
0 • 0 = 0?
∞ • 1 = ∞?
0^∞ = 0?
∞^∞ = ∞?
Or are all of them undefined?
Please could someone explain it to me why. Thank you.
That’s undefined , or the concept of undefined is that there could be many answers and we dk which one to chose , like in infinity x 1 , it could be a 10x1 or a 11x1 , which one is it ? We don’t know , so we say the answer is undefined , there could be an answer , but there are actually so many we don’t know , indeterminate means there is no solution . That is different than many solutions and not being able to tell .
These are all undefined ,
Except maybe 0x0 , maybe that just 0 lol …
sir Give me a example,0/0forms
i like it
0^0 is *not* indeterminate. Many proofs exist and can be found with one cursory search. It turns out that 0^0 = 1. The way to approach values like these with limits does not always yield the correct value. For example:
Ceiling(1) = 1, but
Lim(x --> 1+) Ceiling(x) = 2, since the value "1" will never be reached.
Limits only approach values. If 0^0 were approached, it would be indeterminate, since:
Lim(x --> 0+) 0^x = 0 and
Lim(x --> 0) x^0 = 1
But here, there is no limit. No value is being approached, so there is no indeterminate result.
Your post is misinformation, and it is reported.
OHHH, 1/0 is infinite. I thought it was just like.. idk.. dne. That makes sense.
شكرا لك شرح رائع
Thank you!
You're welcome!
0 ÷ 0 = undefined
😂
LOVE FROM INDIA 🇮🇳
saved me LOL! 😂 thanks
here bec my teacher doesn't know how to explain :0 thanks a ton
Undefined
you speak funny magic man!!
helpful
Thanks for watching :)
Is I A Maron sufficient for calculus IIT JEE???please tell me sir...
I'm really not very familiar with JEE - sorry. Thanks for watching though!
it was very helpful love and respect from pakistan
Happy you enjoyed it!
#undefined
0/0 example of ur's is wrong
Itz correct answer is -1
Cant get what u r trying to say..
I cannot find that rule he is talking about, could someone please send me the link to it? Thank you in advance!❤. And thank you for the video @BriTheMathGuy❤
thank you
Thanks
Wow you saved me. I have two days before I write an exam
Good luck!
Thank you sir
Welcome!
Thank you