The answer is correct but the way I did it was wrong. I have to L'Hopital's but how? Reddit calculus
HTML-код
- Опубликовано: 30 сен 2024
- How do we evaluate the limit of x^tan(x) as x goes to 0+? This is a must-know calculus 1 limit that requires L'Hopital's Rule. Make sure we don't just write 0^0 approaches 1 since 0^0 is a limit indeterminate form.
0^0 approaches 0: • finally 0^0 approaches...
0^0 approaches e: • a 0^0 limit that appro...
This question is from Reddit: / zv0jmxpk9r
-----------------------------
Ask me calculus on Threads: www.threads.ne...
Support this channel and get my calculus notes on Patreon: 👉
/ blackpenredpen
Get the coolest math shirts from my Amazon store 👉 amzn.to/3qBeuw6
-----------------------------
#calculus #bprpcalculus #apcalculus #tutorial #math
0^0->0 ruclips.net/video/BThNFV9f-L0/видео.htmlsi=_PuuOGSjJZqDyLtF
Hey, I know that you don't prefer the ILATE method. But, can you please teach me ILATE? My teachers aren't really accepting the DI method and saying that it is "out of syllabus"😭
Can I use Taylor series with tan(x) to solve for the answer?
ILATE and DI have no contradictions, you can use ILATE with DI, and DI is just a tabular representation, if you write the steps out, DI is acceptable for every single calculus course that allows by parts
This is why people saying 0⁰=1 is an issue. It's situational. A limit of the form 3² is not indeterminate, it's 9. A limit of the form 6-4 is not indeterminate, it's 2. A limit of the form 0⁰ is indeterminate, and you must respect that. Sometimes it will be 1, but not all the time.
No one says that the limit of 0^0 is 1
The algebraic operation 0^0 is an unrelated discussion
@@MikehMike01plenty of people say 0⁰ is 1 without even bringing limits into it. these people are wrong but it's a common belief among non-mathematicians
No matter what you do, people will make mistakes. One just has to teach what indeterminate limiting forms are, and expect students to actually learn it. Again, no matter how good a teacher is, some students will still make mistakes.
@@psymar "plenty of people say 0⁰ is 1 without even bringing limits into it. these people are wrong but it's a common belief among non-mathematicians"
It's also a pretty common belief among mathematicians. In situations where exponentiation is _discrete_ (where exponents _mean_ repeated multiplication), 0^0 = 1 is correct. This does not mean that the _limiting form_ 0^0 should be evaluated as one. We have to remember how limits don't have to match the value of a function at the point in question.
@@psymar yes, 0^0 (not a limit) should be defined as 1. That has nothing to do with the limit of 0^0, which is indeterminate.
Math is my love language
In reality, if you want to know the limit of any function approaching 0, you just have to sub in something like 0.000000001 and you'll be able to easily deduce the answer.
Found the engineer
alr
@@Hextor26 I do actually have a degree in engineering lol
It is meaningful, but situational. Many functions in physics and engineering have their "natural scale", and you have to know it before plugging a number. Your number must be small RELATIVE to that scale; and I'm not even mentioning sometimes this scale doesn't exist (along with the limit). So unfortunately its not foolproof
Can I use Taylor series with tan(x)
If you don't split the limit into a product of limits you would have -sin^2(x)/x -> -sin(2x) through L'Hopitals Rule. Finally, substituting x = 0 leads to the same limit of 0 in the exponent and thus 1 for the limit of the original expression as x appproaches 0 from the right.
Can I use Taylor series with tan(x) to solve for the answer?
What condition is required for lim(x->something)f(x)=f(lim(x->something))?
It requires f to be a continuous function. It's derived from the limit of composition of functions
Continuity. It's essentially the definition of continuity.
Continuity at that something, specifically
lavalier mic goes on your collar
this would free your left hand
It's intentional.
BPRP used to use one of those larger, spherical, desk microphones and would just hold it in his left hand as he was doing the video. He switched to a smaller mic, but wanted to keep holding something in the videos, so he held a pokéball plush. I don't know what happened to the pokéball, but he's still keeping the tradition alive by holding the mic.
Don’t worry. Nothing happened to the pokeball. : )
Thanks a lot from Nuremberg!
I taught students about the cotangent table and graph today!
really want a tee like that for myself but they won't ship to my region //:
How to get the t shirt????in Canada that is
Hi there, can you check the Amazon link in the description and let me know if it works for Canada? Thank you!
It says it doesn't deliver to Canada...
@@shaqshock that’s a bummer. I will have to look into that to see what I can do.
@@bprpcalculusbasics please do thx.... Otherwise I will have to find a place in states.... 😢😢😢
How about this on Spring? blackpenredpen.creator-spring.com/listing/cat-and-indeterminate-forms?product=2
Can I use Taylor series with tan(x)
it makes this question way harder
unless you only take the first term of the expansion, noting that tanx~x
😂😂😂😂❤😡😡
A small shortcut lim_{x-> 0^+} tan(x) = x
Can I use Taylor series with tan(x) to solve for the answer?
you should either write tanx~x or lim tanx/x=1, because x is undefined on the right side
can we just say that in the limit to 0 tanx becomes x?
Can I use Taylor series with tan(x) to solve for the answer?
Like it.
Actually it can be proven that if you have an analytic function to the power of another analytic function, both approaching 0, then the result always approaches 1. (You can even find a published paper about this.) So technically the original method is mathematically valid.
All the counter-examples, where the limit does not approach 1, have one or both functions non-analytic at that 0.
"Actually it can be proven that if you have an analytic function to the power of another analytic function, both approaching 0, then the result always approaches 1."
Yes, but that theorem probably wasn't covered in this calculus class.
The original answer didn't say anything about analytic functions, AND it says (at least in the thumbnail that the limit is equal to 0^0 and then is therefore equal to 1, which is blatantly wrong
@@ThomasTheThermonuclearBomb
Well, you don't need to always name the theorem when you use one. Do you always explicitly write "by Pythagoras' theorem" when you use it, for example?
@@DjVortex-w I didn't say name the theorem. I said to mention the analytic functions. I'd write something like "This is the limit as x approaches 0 of an analytic function to the power of another analytic function, so the limit equals 0" but less wordy. And you can't say 0^0=1, period.
Can I use Taylor series with tan(x) to solve for the answer?
How did we pass from sinx/x to 1?
Standard Calc proof in doing derivatives of sin and cos you have to use sin(x)/x is 1 as X approaches 0. If you proved the trig functions using inverses, then you can bypass the sin(x)/x part of the proof.
Can`t we just do it using L' Hospitals rule again? Because, if we evaluate for x=0, we have a 0/0 situation. And if we take the derivative top and bottom, we just get cos(x)/1 and cos(0)=1.
@@marie-juhanna1281 In order to know the derivative of sin(x) you need to use the definition of the derivative, and to evaluate that limit you need to evaluate the limit of sin(x)/x. So for this limit, we can't use L'hopital's rule without knowing the derivative of sin(x), and we can't know the derivative of sin(x) without first evaluating the limit. It's circular reasoning
@@marie-juhanna1281 Reason you do not use L'H is because you need sin(x)/x limit as part of the (most used) proof of of the derivative of sin(x). The logic is circular.
You wrote that wrong. You need grouping symbols: sin(x)/x.
Yes, lim sinx/x = 1, but lim sinx/n is different. The n cancels out and you're left with six.