Squeeze theorem can’t be applied to lim x -> inf of sin(x) because there aren’t two functions that would “squeeze” in on sin(x) as x goes to infinity. You would need one function that’s an upper bound and one function that’s a lower bound that both approach the same number as x approaches infinity. If you use y=1 as your upper bound and y=-1 as your lower bound, those don’t both approach the same number as x -> inf, so squeeze theorem can’t be applied.
Here's the 2 limits as x goes to infinity we will take a look at:
0:23 | sin(x)
2:30 | sin(x)/x
You’re amazing Thank you
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YOU ARE A LITERAL GOD
thanks so much jake!
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Glad I needed this after you uploaded the video
I hope it helped!
Thank you very much. You are a genius. 👍👍🔝🔝
Nicely explained!
Thank you!
Thanks.....BUDDY😁😁😁
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Thank you so much!Its very clear now!
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Good explanation, thank you.👍
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Thank you 💜
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Great explain bro keep going 💙
I will, I’m glad it was helpful!
the numbers exoressed by the general form 1\(n*n)^(1/2+n*n*i) respect the squeeze theorem...these are special zeta complex numbers
Perfect explanation
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thanku bro...
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I love this, thank you!
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Nice video
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thank youu
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7:42 the moment this clicked for me👍🏼
Nice!
thanks!
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We could also say the same for sin(x) and find that his limit is 1 but his limit dont exist whats the difference
Ok i got it to apply squeeze theorem we need sequences and not number thanks
Squeeze theorem can’t be applied to lim x -> inf of sin(x) because there aren’t two functions that would “squeeze” in on sin(x) as x goes to infinity. You would need one function that’s an upper bound and one function that’s a lower bound that both approach the same number as x approaches infinity. If you use y=1 as your upper bound and y=-1 as your lower bound, those don’t both approach the same number as x -> inf, so squeeze theorem can’t be applied.
Very cool
Thanks!
Solve this x sin(1/x).
Ok i will
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