limit x-&gt;inf of (1+4/x)^2x

Can you work out what the limit of

log((1+4/x)^{2x})

is as x approaches infinity?

(edited 10 years ago)

Reply 2

Felix Felicis

13

Original post by SmOgER

Hello,

lim (1+4/x)^2x when x approaches infinity.

I have a hard time solving this... :frown:

How do you rewrite this to be able to use L'Hopital's rule?

I get something like e^(2(1+4/x) , but it obviously won't work this way : |

Unparseable latex formula:

\displaystyle \lim_{x\to \infty} \left(1+\frac{4}{x}\right)^{2x} = \lim_{x\to\infty} \left(1+\frac{1}{\frac{1}{4} x}\right)^{\frac{1}{4} \cdot x \cdot 8} \overset{t = \frac{1}{4} x}= \lim_{t\to\infty} \left( \left(1 + \frac{1}{t}\right)^{t} \right) \right)^{8}

Now if

\lim_{x\to a} f(x) = L \implies \lim_{x\to a} \Big(f(x)\Big)^{n} = L^n

(edited 10 years ago)

Reply 3

Original post by Felix Felicis

Unparseable latex formula:

\displaystyle \lim_{x\to \infty} \left(1+\frac{4}{x}\right)^{2x} = \lim_{x\to\infty} \left(1+\frac{1}{\frac{1}{4} x}\right)^{\frac{1}{4} \cdot x \cdot 8} \overset{t = \frac{1}{4} x}= \lim_{t\to\infty} \left( \left(1 + \frac{1}{t}\right)^{t} \right) \right)^{8}

Now if

\lim_{x\to a} f(x) = L \implies \lim_{x\to a} \Big(f(x)\Big)^{n} = L^n

I like this method. It's very fast if you are allowed to quote the value of

lim_{t \rightarrow \infty} (1+1/t)^t

Smoger: If you aren't allowed to just quote it, then taking logs will allow you to derive it, as in my original post.

Reply 4

OP

Original post by Felix Felicis

Unparseable latex formula:

\displaystyle \lim_{x\to \infty} \left(1+\frac{4}{x}\right)^{2x} = \lim_{x\to\infty} \left(1+\frac{1}{\frac{1}{4} x}\right)^{\frac{1}{4} \cdot x \cdot 8} \overset{t = \frac{1}{4} x}= \lim_{t\to\infty} \left( \left(1 + \frac{1}{t}\right)^{t} \right) \right)^{8}

Now if

\lim_{x\to a} f(x) = L \implies \lim_{x\to a} \Big(f(x)\Big)^{n} = L^n

Now I should have mentioned it earlier, but the answer is e^8.

I lost you when you introduced t=1/4x

I can't quite understand how this worked out... : |

Reply 5

OP

Original post by theOldBean

Can you work out what the limit of

log((1+4/x)^{2x})

is as x approaches infinity?

Hmm... I have never done examples like that. I can't see inf/inf so I could use L'Hopital's and I'am very much confused with that 2x...

I had lim x->0 (1-4x)^[1/x] example which would simply transform to e^ [(ln(1-4x)) / x], so I guessed I could rewrite this one as well in some similiar way

(edited 10 years ago)

Reply 6

Original post by SmOgER

Hmm... I have never done examples like that. I can't see inf/inf so I could use L'Hopital's and I'am very much confused with this 2x.

Hint 1: You can use L'hopital when you get something tending to 0/0

Hint 2: Multiplying by 2x is the same as dividing by 1/(2x)

Reply 7

OP

Original post by theOldBean

Hint 2: Multiplying by 2x is the same as dividing by 1/(2x)

We have 2x in power, right?
What to do with it?

At the moment I thin I will have to just memorize Felix Felicis method as a formula if I fail to understand it...

(edited 10 years ago)

Reply 8

OP

Ok I'am confused and I think I'am inventing bicycle now... But:

\displaystyle \lim_{x\to\infty} \left(1+\frac{1}{ax}\right)^{c} = \displaystyle \lim_{x\to\infty} \left(1+\frac{1}{ax}\right)^{a*x*b} = e^b[br][br][br][br]

Is this right?
Will it matter if we have + or - in brackets or whatever number?

Dunno if I'am allowed posting links, but all my knowledge on limits up until now was this:
http://www.youtube.com/watch?v=PdSzruR5OeE (including 3 examples on the right)
but this one definitely won't fit into any of these.

Can anyone please explain what's going on? :biggrin:

(edited 10 years ago)

Reply 9

Felix Felicis

13

Original post by SmOgER

Now I should have mentioned it earlier, but the answer is e^8.

I lost you when you introduced t=1/4x

I can't quite understand how this worked out... : |

You can use substitution in limits, not dissimilar to how you've been using substitution in integration. The idea is essentially something like this: if you have continuous functions

f

and

g

and (provided the limit exists), then:

\displaystyle \lim_{x\to a} (f \circ g) (x) = \lim_{x\to g(a)} f(x) = L

There are formal conditions for when you are allowed to use the substitution - I think

g

has to be continuous at

a

and

f

has to be continuous at

g(a)

.

IceKidd

...

Saw you quoted me. :ninja:

It depends on how you define some things tbh - some people prefer to show that the limit converges and define that limit to be

e

; others prefer to define

e^x

by its power series and show that the limit is equal to

e^1

, others like to define

\ln{x}

as the inverse of

e^x

and show it like that, it depends what you take as the definition.

I think my favourite way to do it though is to define

e^x

and

\ln x

as inverses and consider the relative areas under the graph of

f(x) = 1/x

over

[n,n+1], \ n > 0

.

(edited 10 years ago)

Reply 10

OP

Original post by Felix Felicis

then:

\displaystyle \lim_{x\to a} (f \circ g) (x) = \lim_{x\to g(a)} f(x) = L

this is crazy LOL

Original post by SmOgER

\displaystyle \lim_{x\to\infty} \left(1+\frac{1}{ax}\right)^{c} = \displaystyle \lim_{x\to\infty} \left(1+\frac{1}{ax}\right)^{a*x*b} = e^b[br]

I'd love to have some comments on this formula and how it depends on what's inside the brackets (can we use it with any expression in brackets? what if ax is in numerator?)

(edited 10 years ago)

Reply 11

Felix Felicis

13

Original post by SmOgER

this is crazy LOL

The notation may be a bit daunting/ difficult to get your head around at first. :tongue:

I'd love to have some comments on this formula and how it depends on what's inside the brackets (can we use it with any expression in brackets? what if ax is in numerator?)

Ok, so firstly we have:

\displaystyle \lim_{x\to a} (f\circ g) (x) = \lim_{x\to g(a)} f(x)

There's a small mistake in your limit - if you want the general case, it should be:

\displaystyle \lim_{x\to\infty} \left(1 + \frac{a}{bx}\right)^{cx}

We can re-write it like so:

\displaystyle \lim_{x\to\infty} \left(1 + \frac{1}{\frac{b}{a} x}\right)^{c \cdot \frac{b}{a} x \cdot \frac{a}{b}}

Now, if we let

g(x) = \frac{b}{a} x

and

f(x) = \left(1 + \dfrac{1}{x}\right)^{\frac{ac}{b} x}

then clearly

(f\circ g) (x) = \left(1 + \frac{1}{\frac{b}{a}x}\right)^{cx}

which is the limit we're trying to evaluating.

So we proceed with the substitution

t = \frac{b}{a} x:

\displaystyle \lim_{x\to \infty} \left(1 + \frac{1}{\frac{b}{a} x}\right)^{\frac{a}{b} \cdot \frac{b}{a} x \cdot c} \overset{t=\frac{b}{a}x}= \lim_{t\to\infty} \left( \left(1 + \frac{1}{t}\right)^{t}\right)^{a/b \cdot c}

(edited 10 years ago)

Reply 12