Your Teacher Is WRONG About This Limit⚠️

Рет қаралды 4,516

Күн бұрын

Hey, Number Ninjas! In this video, we'll uncover some misunderstandings around limits such as the limit of sin(1/x). Properties of the math function such as it oscillating, or simply the fact that we can't plug in 0 for x due to dividing by zero aren't conclusive enough to determine whether the limit exists. Teachers and students alike can sometimes trip on these important details about solving limits and by the end of this video, my goal is to help you understand the real reason the limit doesn't exist.
I do cover one case in the video about another domain-restricted function that I use to illustrate that simply lacking continuity at a point doesn't imply there's no limit for that point. Here's the video that further dives into that other example I used and it's again another video where I uncover some misunderstandings about whether x^2 / x is the same as x. Be sure to check it out for more useful details!
• 99% Get This WRONG. It...
Here's yet another video of a common misunderstanding about using L'Hopital's rule that you'll want to definitely if you've ever used L'Hopital's rule to solve this famous squeeze limit problem:
• Why Using L'Hopital's ...
Hope you enjoy this video! Be sure to like the video if it was useful to you and subscribe for more helpful tips on your next exam! Also, if you have any feedback for me, leave a comment below as it would help me better help you all out!
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Пікірлер: 22

@NumberNinjaDave 4 ай бұрын

In case of any confusion, I use the graph of sin x in this video, not sin(1/x), since it’s easier than the original problem. I recommend it as a strategy, especially if you don’t have a calculator! I do cover one case in the video about another domain-restricted function that I use to illustrate that simply lacking continuity at a point doesn't imply there's no limit for that point. Here's the video that further dives into that other example I used and it's again another video where I uncover some misunderstandings about whether x^2 / x is the same as x. Be sure to check it out for more useful details! kzbin.info/www/bejne/rXmZgHupZ5VrnNUsi=gezck... Here's yet another video of a common misunderstanding about using L'Hopital's rule that you'll want to definitely if you've ever used L'Hopital's rule to solve this famous squeeze limit problem: kzbin.info/www/bejne/q3a5dH6nhqetadksi=3rkIz... Hope you enjoy this video! Be sure to like the video if it was useful to you and subscribe for more helpful tips on your next exam! Also, if you have any feedback for me, leave a comment below as it would help me better help you all out! Also, sorry about the random line on the screen after about the 5 minute mark

@kingbeauregard Ай бұрын

Because I love pain, I would try to epsilon-delta this thing and discover I cannot find a way to do it. The oscillations as x approaches 0 are so fast - and importantly, the range of the function remains -1 to 1 - it's impossible for epsilon to approach 0 as delta approaches 0 (at the value x = 0).

@NumberNinjaDave Ай бұрын

@@kingbeauregard you just proved that pain DOES exist in this dojo 🥷

@kingbeauregard Ай бұрын

@@NumberNinjaDave Epsilon-delta really isn't that bad, though I know its reputation. But for these purposes, all I have to do is show that, for any non-zero value for x that you choose - be it 0.1 or 0.00001 - I can find a value that's even closer to 0, such that sin(1/x) = 1.

@kingbeauregard Ай бұрын

@@NumberNinjaDave I do wish they'd at least teach the concept of epsilon-delta better though. Like, imagine a point on a graph. Now imagine a rectangle centered on that point, that is tall enough that the function doesn't touch either the top or bottom edge. Can you shrink that rectangle down to nothing, such that the function never touches the top or bottom edge at any scaling? If so, then the limit exists. (You reserve the right to change the shape of the rectangle as you go, just make sure that both the width and height hit 0 at the same time. Also, you reserve the right to limit yourself to a narrow domain around your point; and if you have a simpler function that "contains" the original one that you can make a shrinking rectangle for, then the original function will be bound by that simpler shrinking rectangle too.) That's the concept. Beyond that, the technique is not too difficult; just figure out how to extract an "x-a" from the epsilon inequality, and any remaining x's need to be replaced by constants (thus making the simpler function).

@kingforgotten9090 3 ай бұрын

Sin(x) does not equal 1 at 3pi/2 it is equal to -1.

@NumberNinjaDave 3 ай бұрын

True

@henryginn7490 Ай бұрын

Doesn't this limit very clearly not exist? It's oscillating like crazy around 0? What would teachers be getting wrong about this?

@NumberNinjaDave Ай бұрын

Sometimes, answers aren’t obvious and still necessitate a mathematical proof of correctness.

@henryginn7490 Ай бұрын

@@NumberNinjaDave Of course it necessitates a proof, but the naive plan of "it oscillates - let's find two subsequences that don't converge to the same thing" just works quite simply without complication. I'm questioning the title. Are there teachers saying this limit exists? Getting the proof wrong somehow? What would a hypothetical teacher be getting wrong about this limit.

@DirectedArt 2 ай бұрын

What i would do is to make a variable change θ=1/x, x->0 => θ->∞ so we have the limit when θ->∞ of sinθ. Now we do know that -1≤sinθ≤1 so we take the limit in all 3 parts of the inequality, and the limit of -1 and 1 is -1 and 1, respectively so the limit oscilates between -1 and 1 so the limit diverges

@NumberNinjaDave 2 ай бұрын

👏👏👏🥷

@ezraoctopus 4 ай бұрын

Can't you just say that as x goes to 0, the argument goes to infinity or negative infinity, and since sin oscilates as it approaches infinity and negative infinity, the limit doesn't exist?

@NumberNinjaDave 4 ай бұрын

@@ezraoctopus It’s not wrong to do that but we need to show work to prove it.

@NumberNinjaDave 4 ай бұрын

Starting next week, Saturday video uploads will go back to 6AM MST instead of 6PM since it looks like you guys prefer that! Keep crushing it, ninjas!