Question

Find the limit of (1-3/x)^2x as x approaches positive infinity

Answer 1

Is anyone here taking up calculus?

Answer 2

Combine 1-(3)/(x) into a single expression
\[\lim_{x \rightarrow \infty}(\frac{x-3}{x})\]


The value of \[\lim_{x \rightarrow \infty}(\frac{x-3}{x})\] is 1

Answer 3

have you seen

\[\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}=e^x\]

Answer 4

Why is the answer 1?

Answer 5

Mr. Zarkon is good so you can do too write like this [(1 + -3/x)^x]^2

Answer 6

by the way that equation is still raised to 2x

Answer 7

ignore my answer follow zarkon's one @Naomi22 yes i didn't notice there's power 2x

Answer 8

I prefer Lord Zarkon, King Zarkon or Dr. Zarkon... as appose to Mr. Zarkon :)

Answer 9

Ok yes sir.. thank you.

Answer 10

alright. so how can I write this in terms of e

Answer 11

combine what WonHungLo and I wrote

Answer 12

e^[(1-3/x)^2x] like that?

Answer 13

no

Answer 14

oh yeah where did the limit go?

Answer 15

you e^(limit)

Answer 16

oh ok

Answer 17

\[\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{b x}=e^{ab}\]

Answer 18

is this right?
e^lim[(1-3/x)^2x]

Answer 19

no

Answer 20

nvm ,use zarkon's way

Answer 21

I'll just directly copy zarkon's equation on paper then plug in my values. It's easier that way

Answer 22

what do you get for your final answer?

Answer 23

could you solve it by the way?

Answer 24

wait I'm getting 0 x infinity

Answer 25

I can...can you?...what do you get?

use the formula I gave you

Answer 26

we're told to use L'hopital's rule on this one

Answer 27

but using your formula it's e^-6

Answer 28

yes...that is the answer

Answer 29

great thanks. but I still have to figure out a way to do this applying L'hopital's rule

Answer 30

if you want to calculate it with out the formula then

\[\lim_{x\to\infty}\left(1-\frac{3}{x}\right)^{2x}\]

let \[y=\left(1-\frac{3}{x}\right)^{2x}\]

then \[\ln(y)=2x\ln\left(1-\frac{3}{x}\right)\]


which can be written as

\[\ln(y)=2\frac{\ln\left(1-\frac{3}{x}\right)}{\frac{1}{x}}\]

now take the limit using L'Hospitals rule

Answer 31

alright thank you very much Lord Zarkon. Your name fits you just fine :D

Answer 32

no problem