I
iTeachChemā€¢7d ago
Opt

Limits

Let's see if I remember how to type this out. $\lim_{x \to 0}\frac{cos(sinx)-cosx}{x^4}$ I tried series expansion, and got somewhere, but it didn't work out in the end.
38 Replies
TeXit
TeXitā€¢7d ago
Opt
No description
iTeachChem Helper
iTeachChem Helperā€¢7d ago
@Apu
iTeachChem Helper
iTeachChem Helperā€¢7d ago
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.
SirLancelotDuLac
SirLancelotDuLacā€¢7d ago
Series expansion works na? In this? Just make sure you use sine ka series expansion also. Like $(x-\frac{x^{3}}{6})=sin(x)$ le ke sine and cosine ka series expansion
TeXit
TeXitā€¢7d ago
SirLancelotDuLac
No description
Nimboi
Nimboiā€¢7d ago
you're suggesting series expansion of sin inside a cos and then series expanding that expansion in the argument of cos?
Opt
OptOPā€¢7d ago
Oh, I just expanded the cosines
SirLancelotDuLac
SirLancelotDuLacā€¢7d ago
That would ignore several x^4 terms that come by sine wala lagane mein ig.
flower
flowerā€¢7d ago
I have the question in my module šŸ˜†šŸ˜†šŸ˜†
Opt
OptOPā€¢7d ago
Yeah, I got those zeroes in the denominator and went nah
SirLancelotDuLac
SirLancelotDuLacā€¢7d ago
First expand by cosine and then apply sine in my opinion.
flower
flowerā€¢7d ago
Achcha so like we first assume like sinx to just not exist in the cos, write the cos ka expansion and then put sinx everywhere where it should
Opt
OptOPā€¢7d ago
Ye I guess that should work
SirLancelotDuLac
SirLancelotDuLacā€¢7d ago
Ye and then apply sine ka expansion. Even after that its still painful tho.
flower
flowerā€¢7d ago
AYEIN NAHH ATP JUST SKIP THE QUESTION šŸ˜­šŸ˜­
Opt
OptOPā€¢7d ago
It is. Three terms at least in each expansion, because xā“ in denom That's what I did lol.
SirLancelotDuLac
SirLancelotDuLacā€¢7d ago
Yeah. In paper just have a hindsight what to ignore ig.
Opt
OptOPā€¢7d ago
And four mistakes. So 76 in maths today
BlindSniper (BS)
BlindSniper (BS)ā€¢7d ago
not that difficult tho send paper
Opt
OptOPā€¢7d ago
I will. I made silly errors lol
BlindSniper (BS)
BlindSniper (BS)ā€¢7d ago
76 is pretty wild
Opt
OptOPā€¢7d ago
It was easy
BlindSniper (BS)
BlindSniper (BS)ā€¢7d ago
I'll judge
Dexter
Dexterā€¢7d ago
2/3 @Opt ? no 1/3 sorry eh?
Opt
OptOPā€¢7d ago
1/6 diya hai
Dexter
Dexterā€¢7d ago
naw wait
Opt
OptOPā€¢7d ago
It's in that paper I just sent
BlindSniper (BS)
BlindSniper (BS)ā€¢7d ago
oh it actually is pretty easy damn
Dexter
Dexterā€¢7d ago
yeah 1,-1 krna hai bas
BlindSniper (BS)
BlindSniper (BS)ā€¢7d ago
was talking about the paper
Dexter
Dexterā€¢7d ago
oh sorry šŸ˜­ got it lemme send the soln
Opt
OptOPā€¢7d ago
Ok, yeah, I think I got this.
Dexter
Dexterā€¢7d ago
@Opt sorry for the mess
No description
Dexter
Dexterā€¢7d ago
just standard limits and L hopital
Opt
OptOPā€¢7d ago
Understood. Series expansion worked out for me.
Dexter
Dexterā€¢7d ago
good stuff man
Opt
OptOPā€¢7d ago
Aight, imma close this. +solved @Dexter @BlindSniper (BS) @SirLancelotDuLac
iTeachChem Helper
iTeachChem Helperā€¢7d ago
Post locked and archived successfully!
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