Swartwoodprep
Region: US
Tuesday 06 October 2026 19:43:06 GMT
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YNWniceshot :
yo i couldn't care less abt the math how did you do that dotted line
2026-10-07 04:18:25
256
examinermo :
Just plug in a couple numbers on each section and you’ll see the square root section always gives you a bigger number. I realized it once he squared both sides. So the answer is the first infinity is bigger the limit goes to infinity and x approaches infinity. And beyond.
2026-10-08 03:53:03
0
emeraldzs :
one thing I hate about higher level math is that it feels so experimental, "try it out and see if it works". Solutions just feel so unintuitively formed-more stumbled upon and you have to build a library of techniques rather than principle which is much more satisfying and free in problem solving
2026-10-07 05:07:12
33
Amahle.Thwala :
i feel like we have bigger problems 🥱😮💨 like that dotted line in a single motion 🫡😳 like how ?
2026-10-06 21:19:20
79
nicholas17k :
showed me RIGHT after the limit test 😭
2026-10-08 21:34:04
1
Albo76 :
Can you explain why the limit would not be 0?
2026-10-06 21:08:27
9
slow bear :
When you squared x and then took the root you basically took absolute value of x and |x| is not equal to x so isn’t that wrong? Because you divided other values by x and other one by |x| so if x is negative for example those will be different numbers
2026-10-08 17:59:31
2
Ho Vi596 :
His chalk can fart 💨. Awesome!
2026-10-08 22:33:38
3
Qmltu :
Im sorry but why are you writing so aggressively😭
2026-10-07 21:31:44
3
Sedqi :
So if \sqrt{x^2+x} \sim x as x \to \infty, why can't I just plug in x for \sqrt{x^2+x} directly in the expression? And how do we figure out the first correction term we need to keep to evaluate \lim_{x\to\infty}x\left(\sqrt{x^2+x}-x\right) without using L'Hôpital's rule? Also, does asymptotic equivalence still hold when you subtract two functions that are close to each other?
2026-10-09 07:06:09
0
michaelrussell020 :
So if both infinities lose, you get 0? [Tears of joy]
2026-10-08 10:48:11
0
Siobansny :
Could you get the same through completing the square within the root on the first line so sqrt((x+1/2)^2-1/4) - x? For large values of x that -1/4 in the root becomes more and more negligible. I’m just trying to think of a rule or practice that would allow us to use that process.
2026-10-07 20:08:35
1
Donald Butler :
You only need to determine which is bigger. It should be obvious that x^2+x > x^2, so the limit is +infinity.
2026-10-08 00:11:25
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Tingpeptide :
you could show that the whole expression is less than 1/2 and that 1/2 is the least upper bound
2026-10-07 06:48:35
1
🎟️ :
really helpful🙏
2026-10-06 20:21:13
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Ilyes :
greate 👏👏👏
2026-10-06 20:06:07
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Siniestro :
Great!!!
2026-10-07 22:17:10
1
Lester :
Excellent!
2026-10-07 13:13:13
1
pahl.gabriel :
Really well done!
2026-10-07 01:47:31
2
cosmic_evil :
smooth with the chalk
2026-10-07 01:59:27
1
crow :
let me js hop on the game
2026-10-07 07:19:24
0
Izzy :
plug in 1000 for x and call it a day
2026-10-06 23:05:29
3
م. :
If he has long hair he can prolly invent maths😭🙏
2026-10-07 14:15:59
0
daddy1db76151 :
so sexy
2026-10-06 21:53:49
0
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