Indeterminate Forms

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information. Viewers should always verify the information provided in this video by consulting other reliable sources.

[carly]

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

[satanlucifer]

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

[meganevans]

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!!

[alaayousef]

I really don't know what to say, but THANK YOU THAT'S REALLY HELPFUL!!!😭❤

[Anima]

Have a great day too and thank you for your guidance

[aoi]

This was the cleanest explanation for solving indeterminate forms I've EVER seen in my life, you're a lifesaver dude.

[Romero]

thought this was patrick JMT when i clicked on it

[santialterman]

perfectly consice and informative. Thank you!

[lidyakumar]

love this! this was clear and to the point!

[kyonhihorahipadhai]

Why didn't I find this sooner? 😩
Fantastically explained 👍🏼

[eyadaboelgoud]

man keep going . i really love this orginaize you do. it helps alot. thank you so much

[jamesperalta]

This is the best video on this topic, no cap!

[vivekvana]

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.

[anamaed.gamban]

it very helpful for me for my reports in calculus 2, thank you so much

[agrimpuriya]

Damn. You're a lifesaver. Thanks. Subscribed.

[AlbertTheGamer-gksn]

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.

[cubekaraja]

But some says that 1/0 is not defined and it is not infinity

[minyo]

wow this was so easy to understand. Thank you!!

[AditiSingh-hefq]

is there any other way of simplifying the 3rd type of indeterminants?

[sahintaufika]

Thank you so much sir... I am eager to know about the 1/0, this indeterminate form