/sci/ - Science & Math » Thread #11525257

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Explain this.

Anonymous Thu 02 Apr 07:31:11 2020 No.11525257 View Reply Original Report

Quoted By: >>11525282 >>11525283 >>11525285 >>11526289

I am a newbie in Precalculus and I don't get it why there are so many ways to do calculations in limits like Direct subtraction, Factorial and this one?

Watch this in Khan Academy
Link: https://youtu.be/gWSDDopD9sk

Help me to learn calculus I am so bad since coronavirus told us to stay at home.

Anony?m??o?us??

Anony?m??o?us?? Thu 02 Apr 2020 07:45:13 No.11525273 Report

Quoted By:

my god math is so sad and retarded.

https://www.wolframalpha.com/input/?i=lim%28x-%3Einf%29+%28%28x%5E2%29+%2B3%29%2Fx%5E2

implicitly it's saying x = infinity at some impossible point that doesn't actually exist, so it turns into $\frac{\infty}{\infty} = 1$ even though inf/inf is undefined.

what a sad state math is in. why is all of this made up fantasy nignog mystical chinese dragon math even considered.

do something better with your time m8.

A?no??n?y?mou?s

A?no??n?y?mou?s Thu 02 Apr 2020 07:48:24 No.11525279 Report

Quoted By: >>11525286

$\displaystyle \lim_{x \rightarrow \infty } \dfrac{x^2+3}{x^3} = \lim_{x \rightarrow \infty } \frac{1}{x}+\frac{3}{x^3}=0+0=0 \\ \displaystyle \lim_{x \rightarrow \infty } \dfrac{3x^2+x}{4x^2-5} = \lim_{x \rightarrow \infty } \frac{3+1/x}{4-5/x^2}+\frac{3+0}{4-0x^3} = \dfrac{3}{4} $

A?no?ny??m?o?us

A?no?ny??m?o?us Thu 02 Apr 2020 07:49:15 No.11525281 Report

Quoted By:

>x=?
lol

An?o?n??y??mous

An?o?n??y??mous Thu 02 Apr 2020 07:49:40 No.11525282 Report

Quoted By:

>>11525257
$\displaystyle \lim_{x \rightarrow \infty } \dfrac{x^2+3}{x^3} = \lim_{x \rightarrow \infty } \frac{1}{x}+\frac{3}{x^3}=0+0=0 \\ \displaystyle \lim_{x \rightarrow \infty } \dfrac{3x^2+x}{4x^2-5} = \lim_{x \rightarrow \infty } \frac{3+1/x}{4-5/x^2}+\frac{3+0}{4-0} = \dfrac{3}{4} $

Anonymous

Anonymous Thu 02 Apr 2020 07:50:26 No.11525283 Report

Quoted By:

>>11525257
lee hospitals rule

An?on?y??mo?u?s

An?on?y??mo?u?s Thu 02 Apr 2020 07:51:25 No.11525285 Report

Quoted By: >>11525289 >>11525292

Anonymous

Anonymous Thu 02 Apr 2020 07:51:29 No.11525286 Report

Quoted By: >>11525291

>>11525279
> 3+1
>X+x^3

What the fuck is going on?

Anonymous

Anonymous Thu 02 Apr 2020 07:52:34 No.11525289 Report

Quoted By: >>11525293

>>11525285
Explain it.

Ano?n?ym?o?u?s?

Ano?n?ym?o?u?s? Thu 02 Apr 2020 07:53:14 No.11525291 Report

Quoted By:

>>11525286
(20 + 30) / 5 = 20/5 + 30/5

Anonymous

Anonymous Thu 02 Apr 2020 07:54:06 No.11525292 Report

Quoted By: >>11525295 >>11525301

>>11525285
> 4-5 =4

Explain?

Ano?n?ym?ou??s?

Ano?n?ym?ou??s? Thu 02 Apr 2020 07:54:32 No.11525293 Report

Quoted By:

>>11525289
(1/x^2)/(1/x^2) = 1

An?o??n??ymou?s

An?o??n??ymou?s Thu 02 Apr 2020 07:55:39 No.11525295 Report

Quoted By:

>>11525292
you won't make it

Anonymous

Anonymous Thu 02 Apr 2020 07:56:20 No.11525298 Report

Quoted By: >>11525309

Use an \epsilon - \delta argument. The function is a polynomial so it shouldn't be too hard proving it.

Anonymous

Anonymous Thu 02 Apr 2020 07:56:43 No.11525301 Report

Quoted By: >>11525309

>>11525292
>> 4-5 =4
you can't be serious, it's 4-5/(x^2) as x gets fucking huge, so it's 4 b/c 5 is tiny compared to huge ass x^2.

Anonymous

Anonymous Thu 02 Apr 2020 08:01:57 No.11525309 Report

Quoted By:

>>11525298
Sorry, I didn't really do good in math, the last sem I failed my algebra class.

And also the education system here is very poor.

>>11525298
What do you mean? I am just learning.

>>11525301
You mean PEMDAS?

Anonymous

Anonymous Thu 02 Apr 2020 16:44:35 No.11526289 Report

Quoted By:

>>11525257
When working with limits at your level, direct substitution works for any function with no discontinuities and not examined at infinity. everything is nice and smooth there's no jumps, asymptotes etc. Everything else is just a way to find the limit when the function doesn't exist at that point, or is a limit at infinity.

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