/sci/ - Science & Math » Thread #10924862

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Limits

Anonymous Wed 28 Aug 09:45:42 2019 No.10924862 View Reply Original Report

Quoted By: >>10924871 >>10924872 >>10924873 >>10924877 >>10924922 >>10924943 >>10924956 >>10925588

Why is this wrong?
I know the answer is supposed to be $e^-3$ and I'm supposed to do it differently, but how is this wrong?

Anonymous

Anonymous Wed 28 Aug 2019 09:53:55 No.10924871 Report

Quoted By:

>>10924862
I don't see how this is supposed to equal e^-3. It's e^-2.

Anonymous

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Anonymous Wed 28 Aug 2019 09:54:31 No.10924872 Report

Quoted By: >>10926592

>>10924862

Anonymous

Anonymous Wed 28 Aug 2019 09:54:54 No.10924873 Report

Quoted By: >>10925701

>>10924862
1^? is not equal to 1. It's an indeterminate form.

Anonymous

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Anonymous Wed 28 Aug 2019 09:57:51 No.10924877 Report

Quoted By: >>10924903 >>10924907 >>10924961

>>10924862
By your logic e=1. You are retarded, try reading a book for once.

Anonymous

Anonymous Wed 28 Aug 2019 10:18:16 No.10924903 Report

Quoted By:

>>10924877
This

Anonymous

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Anonymous Wed 28 Aug 2019 10:19:08 No.10924907 Report

Quoted By: >>10924914 >>10925001 >>10925165

>>10924877
I know you think you're smart, but how are n and n+5 the same?

Anonymous

Anonymous Wed 28 Aug 2019 10:28:22 No.10924914 Report

Quoted By: >>10925165

>>10924907
Anon, if any two of lim a, lim b or lim a/b exist and provided lim b != 0 then the third one also exists.
You can just use that.

Anonymous

Anonymous Wed 28 Aug 2019 10:32:41 No.10924922 Report

Quoted By: >>10924948

>>10924862
You need to use the exponent rule: $\quad \:a^n=e^{\ln \left(a^n\right)}=e^{n\cdot \ln \left(a\right)}$

$\left(\frac{n+2}{n+5}\right)^n=e^{n\ln \left(\frac{n+2}{n+5}\right)} =\lim _{n\to \infty \:}\left(e^{n\ln \left(\frac{n+2}{n+5}\right)}\right)$

then the $n\ln \left(\frac{n+2}{n+5}\right)$ part is just -3 by L'Hopital's Rule. So $e^-3$ .

Anonymous

Anonymous Wed 28 Aug 2019 10:47:19 No.10924943 Report

Quoted By:

>>10924862
Because even though the expression in the parentheses is very close to 1, you raise it to a huge power.
For example consider 1.000001 = (1+1/10^6).
Raise that to the 10^6 (multiply it by itself 1 million times) and you'll get 2.718280469319376883819799708454356392751645026682507712940... which is close to e.

Anonymous

Anonymous Wed 28 Aug 2019 10:52:33 No.10924948 Report

Quoted By: >>10924969

>>10924922
Right, I did this and it worked. But can I simply plug in 999999999999 instead of n? It's the same basically.

Anonymous

Anonymous Wed 28 Aug 2019 10:55:42 No.10924956 Report

Quoted By: >>10924961 >>10926152 >>10927655 >>10927731

>>10924862
$(\frac{n+2}{n+5})^n = \frac{(1+\frac{2}{n})^n}{(1+\frac{5}{n})^n} \to \frac{e^2}{e^5} = e^{-3}$

Anonymous

Anonymous Wed 28 Aug 2019 10:58:21 No.10924961 Report

Quoted By:

>>10924956
Shiiiet, >>10924877 was right, I am retarded.

Anonymous

Anonymous Wed 28 Aug 2019 11:02:14 No.10924969 Report

Quoted By: >>10924979 >>10926428

>>10924948
No because that just gives you 1. Use LHR - take the derivative of the top and the derivative of the bottom (don't use quotient rule). Continue taking derivatives until you get something that isn't $0/0$ or $\infty / \infty$ . In your example, you're just left with -3.

Anonymous

Anonymous Wed 28 Aug 2019 11:09:07 No.10924974 Report

Quoted By: >>10924979

You can use a Taylor series because (sorry no math mode on mobile)
(n+2)/(n+5)=((n+2)/n)*(1/(1+5/n))=(1+2/n)*(1-5/n)+O(n^2)=1-3/n+O(n^2)

Anonymous

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Anonymous Wed 28 Aug 2019 11:10:46 No.10924979 Report

Quoted By: >>10924994 >>10925712

>>10924974
>>10924969
Poor kid is probably in his first calc class. Give him a break with this L'Hopital and Taylor Series shit lmao.

Anonymous

Anonymous Wed 28 Aug 2019 11:17:32 No.10924994 Report

Quoted By:

>>10924979
Taylor expansions are the most often used tool I learned in undergrad, seriously. Good to learn these early.

t. Physicist

Anonymous

Anonymous Wed 28 Aug 2019 11:20:42 No.10925001 Report

Quoted By:

>>10924907
Not OP but they are as n grows without bound.

Anonymous

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Anonymous Wed 28 Aug 2019 13:16:40 No.10925165 Report

Quoted By:

>>10924907
See >>10924914

Anonymous

Anonymous Wed 28 Aug 2019 17:58:08 No.10925588 Report

Quoted By:

>>10924862
You cant just take the limit of (1 + 2/n)/(1 + 5/n) independently of the exponent. This is literally the whole reason why (1 + 1/n)^n ‘s limit is e.

Anonymous

Anonymous Wed 28 Aug 2019 18:46:53 No.10925701 Report

Quoted By:

>>10924873
this

Anonymous

Anonymous Wed 28 Aug 2019 18:50:24 No.10925712 Report

Quoted By:

>>10924979
L’Hopital is the most kino result in calc I. he needs to learn or perish

Anonymous

Anonymous Wed 28 Aug 2019 21:27:46 No.10926152 Report

Quoted By:

>>10924956
/thread

Anonymous

Anonymous Wed 28 Aug 2019 23:27:00 No.10926428 Report

Quoted By: >>10927134

>>10924969
Er, what do you consider to be the top?

Anonymous

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Anonymous Thu 29 Aug 2019 00:43:43 No.10926592 Report

Quoted By:

>>10924872
>tfw you will never get to see the lohan career we all deserved
ITS NOT FAIR BROS

Anonymous

Anonymous Thu 29 Aug 2019 04:26:26 No.10927134 Report

Quoted By:

>>10926428
I was referring to using LHR on $\lim _{n\to \infty \:}\left(e^{n\ln \left(\frac{n+2}{n+5}\right)}\right)$ . You would rewrite the fraction in the ln() part as $\frac{(n+2)(n+5)}{(1/n)}$ . You would then take derivatives of the top and bottom part separately.

Anonymous

Anonymous Thu 29 Aug 2019 11:31:25 No.10927655 Report

Quoted By:

>>10924956
Very good anon. /thread

Anonymous

Anonymous Thu 29 Aug 2019 12:44:16 No.10927731 Report

Quoted By: >>10927735 >>10927851

>>10924956
This is clearly how you're mean to do it, it's a lot cleaner than all this power series or l'Hopital shit.

Anonymous

Anonymous Thu 29 Aug 2019 12:47:29 No.10927735 Report

Quoted By:

>>10927731
...it's implicit

Anonymous

Anonymous Thu 29 Aug 2019 14:17:06 No.10927851 Report

Quoted By:

>>10927731
it's a limit, you're not suppose to solve it algebraically

Anonymous

Anonymous Thu 29 Aug 2019 14:55:46 No.10927921 Report

Quoted By:

English is not my native language sorry
Think about huge values, like (10^100)/(10^100-2). Does the -2 really make much of a difference? Now try for larger and larger numbers and see how it approximates to 1. A limit is basically the number that the expression aproximates to without ever stopping.
In the case of 1^infinite; 1 ^ any number no mater how big is always one, but the expression is NEVER 1. A number very near 1 (eg 1.000000000001) to the power of something, makes the expression increase. The aproximation of 1 to the aproximation of larger and larger numbers without ever stopping aproximating tends to e

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