Find the mistake first, then do this correctly!

Рет қаралды 5,227

Күн бұрын

Пікірлер: 25

@cyrusyeung8096 4 ай бұрын

For polar coordinate: x sin(xy) / (x² + y²) = r cosθ sin(r²sinθcosθ) / r² = r sinθ cos²θ sin(r²sinθcosθ) / (r²sinθcosθ) As r approaches 0, r²sinθcosθ approaches 0 (as |sinθcosθ| ≤ 1 < +∞) Therefore, sin(r²sinθcosθ)/(r²sinθcosθ) approaches 1 Note that |sinθcos²θ| ≤ 1 < +∞ Hence, the limit xsin(xy)/(x² + y²) = (0)(sinθcos²θ)(1) = 0

@bprpcalculusbasics 4 ай бұрын

Excellent!

@EmmanuelGiouvanopoulos 4 ай бұрын

he got an excellent from the professor himself! I constructively envy you, my friend!

@brycenbg9372 4 ай бұрын

wait, where is the whiteboard? 😂

@vr_9171 4 ай бұрын

Step 3 is possible but inaccurately justified. Just consider the fact that x² = |x|² and you are good to go.

@monzurrahman8307 4 ай бұрын

But then you would end up with |sin(xy)|/|x| rather |x||sin(xy)|. You can still proceed by multiplying by |y|/|y|, but step 3 is actually wrong

@vr_9171 4 ай бұрын

@@monzurrahman8307 Yeah I meant possible but by another way and a different result. After this step you may multiply up and down by xy and use sin(xy)/xy → 1 to conclude

@bitoty9357 4 ай бұрын

new white board?

@cyrusyeung8096 4 ай бұрын

Bprp said he was in Taiwan, and he bought a new whiteboard from 蝦皮（an e-commerce platform like eBay）

@kiransabapathy6347 4 ай бұрын

Can you help me solve the integral (secant ^x (x) dx.

@DaiShuryoTechnus 4 ай бұрын

How is y^2 gone?

@monzurrahman8307 4 ай бұрын

If you continue with the original method: 3) x² = |x|², so |x||sin(xy)|/x² = |sin(xy)|/|x|. Assume y ≠ 0, we can multiply by |y|/|y| and bring the denominator inside to get |y| |sin(xy)/(xy)|, which goes to 0. Finally, if y = 0, then the original expression immediately collapses to 0, as x ≠ 0.

@bprpcalculusbasics 4 ай бұрын

Why didn’t I think of that?! Thanks.

@bprpcalculusbasics 4 ай бұрын

Check out the 6 ways of evaluating the limit of a multi-variable function: kzbin.info/www/bejne/fHuQoJmGi75ohposi=sn5NrvmLhM-VyVNO

@EmmanuelGiouvanopoulos 4 ай бұрын

wait, why paper and no board?

@nvapisces7011 4 ай бұрын

I would have used |x|

@jackkalver4644 4 ай бұрын

Multiplying by x^2 was the mistake. Since x≈0, |x||sin(xy)|/x^2>=|x||sin(xy)|.

@Yasser-xl7rt 12 күн бұрын

Step number 3

@ymj5161 4 ай бұрын

最关键的一点还是limx→0, (sinx)/x=1

@rootroot-pi 4 ай бұрын

NEW BOARD!!!

@rootroot-pi 4 ай бұрын

where is the truck

@sampreetkaur3447 4 ай бұрын

okay , this is very random , but I wish you could teach me how to differentiate Schrödinger's equation 🥺. I am a high school student and i learn your calculus to solve calculus based physics😭. I WISH YOU COULDDDDDDDDDDDDDDD PWEASESSSSSSSSSS.