Question

How should I begin to calculate the following limit?
\[\lim_{x \rightarrow +\infty} (\frac{x+4}{x+2})^x\]
Does it have relation with the orders of both numerator and denominator? So the answer to the question should be 1? Or is it wrong?

Answer 1

\[\lim_{x \rightarrow +\infty} (\frac{x+4}{x+2})^x\]

Answer 2

take the log. then compute the limit, then exponentiate. by which i mean if your answer to the limit of the log is 3, your actual answer will be
\[e^3\]

Answer 3

so step one is
\[x\ln(\frac{x+4}{x+2})\] which looks like infinity times 0

Answer 4

step two rewrite as
\[\frac{\ln(\frac{x+4}{x+2})}{\frac{1}{x}}\]

Answer 5

standard trick when you are faced with
\[0\times \infty\]

Answer 6

now use l'hopital to find your limit

Answer 7

the final answer is \[e^2\]

Answer 8

not so easy now is it? but that is ok just need the derivative of top and bottom
derivative of
\[\frac{1}{x}\] is
\[-\frac{1}{x^2}\]

Answer 9

derivative of
\[\ln(\frac{x+4}{x+2})\] is
\[-\frac{2}{x^2+6x+8}\]

Answer 10

so bring up the denominator and you get
\[\frac{2x^2}{x^2+6x+8}\]
take the limit as x goes to infinity with your eyeballs and get limit is 2
so answer is... what zarkon said!

Answer 11

all the gory details are there so i hope it is clear, more or less

Answer 12

Thank you very much, satellite73! Very clear explanation!

Answer 13

if you know that

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

then you can do the following

\[\lim_{x \rightarrow +\infty}\left (\frac{x+4}{x+2}\right)^{x}\]

\[=\lim_{x \rightarrow +\infty}\left (\frac{x+2+2}{x+2}\right)^{x}\]

\[=\lim_{x \rightarrow +\infty}\left (\frac{x+2}{x+2}+\frac{2}{x+2}\right)^{x}\]

\[=\lim_{x \rightarrow +\infty}\left (1+\frac{2}{x+2}\right)^{x}\]

\[=\lim_{x \rightarrow +\infty}\left (1+\frac{2}{x+2}\right)^{x+2-2}\]

let \[w=x+2 \text{ then as }x\to\infty, w\to\infty\]

\[=\lim_{w \rightarrow +\infty}\left (1+\frac{2}{w}\right)^{w-2}\]

\[=\lim_{w \rightarrow +\infty}\left (1+\frac{2}{w}\right)^{w}\left (1+\frac{2}{w}\right)^{-2}\]

\[=e^2\cdot 1=e^2\]

Answer 14

wow. i guess you can do that. it is good to know this way too.

Answer 15

satellite73 technique is needed to verify that

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

Answer 16

but once you have it ...you might as well use it ;)

Answer 17

Yeah, thanks Zarkon for this another way of solving it!

Answer 18

@zarkon no, not really. in fact that
\[\lim_{n\rightarrow \infty}(1+\frac{x}{n})^{n}\] is as good a definition of
\[e^x\] as any

Answer 19

if that is your definition of e^x then sure

otherwise it needs some kind of proof

Answer 20

gj you guys
i was kindof thinking of zarkon's way before i scrolled down and see his answer
i like his answer best :)

Answer 21

no you don't! you are just trying to be mean to me. i bet you $1,000,000,000 that you would have done just what i did. take the log etc. so don't try to play me young lady. i have a whole stalk of sensei

Answer 22

lol