please help

evaluate lim as x goes to infinity (1+(a/x))^x

Printable View

- May 22nd 2008, 12:58 AMGigabytelimits
please help

evaluate lim as x goes to infinity (1+(a/x))^x - May 22nd 2008, 01:07 AMIsomorphism
- May 22nd 2008, 01:07 AMMoo
Hello,

$\displaystyle \lim_{x \to \infty} {\color{red}\ln\left[\left(1+\frac ax\right)^x\right]}=\lim_{x \to \infty} x \ln \left(1+\frac ax\right)$

Change the variable : $\displaystyle t=\frac ax \implies x=\frac at$

When $\displaystyle x \to \infty$, $\displaystyle t \to 0$

$\displaystyle \lim_{x \to \infty} x \ln \left(1+\frac ax\right)=\lim_{t \to 0} \ln (1+t) \cdot \frac at=\lim_{t \to 0} a \cdot \frac{\ln(1+t)}{t}=a \cdot \lim_{t \to 0} \frac{\ln(1+t)-\ln(1)}{t}$

But $\displaystyle \lim_{t \to 0} \frac{\ln(1+t)-\ln(1)}{t}$ is, according to the definition of the derivative number, $\displaystyle \left(\ln(1+t)\right)'_{t=0}=\frac{1}{1+t} \bigg|_{t=0}=1$

Therefore, the limit is $\displaystyle \lim_{x \to \infty} {\color{red}\ln\left[\left(1+\frac ax\right)^x\right]}=a \cdot 1=\boxed{\color{red}a}$

But $\displaystyle \left(1+\frac ax\right)^x=e^{\color{red}\ln\left[\left(1+\frac ax\right)^x\right]}$

---> $\displaystyle \boxed{\lim_{x \to \infty} \left(1+\frac ax\right)^x=e^a}$ - May 22nd 2008, 01:44 AMGigabyte
thanks heaps for that bru.

Is 1 line wrong though.

lim as t tends to 0

Isn't it supposed to be:

ln(1+t)^(a/t)

Not:

ln(1+t)*(a/t) as you wrote it?? - May 22nd 2008, 01:47 AMMoo
- May 22nd 2008, 02:06 AMGigabyte
yip that makes sense. Sorry one of those Rookie errors.

Why do you get?

(ln(1+t)-ln(1))/t

When you take (a) out the front of the limit. You take (a) out the front cos it's a constant right?

Then after that i kinda didn't get the next part.

"But is, according to the definition of the derivative number, "

Could you please explain that because i think i'm getting. (Rofl) - May 22nd 2008, 02:08 AMMoo
$\displaystyle \ln(1)=0$, so I could add it, right ? :)

Quote:

When you take (a) out the front of the limit. You take (a) out the front cos it's a constant right?

Quote:

Then after that i kinda didn't get the next part.

"But is, according to the definition of the derivative number, "

Could you please explain that because i think i'm getting. (Rofl)

- May 22nd 2008, 02:16 AMIsomorphism
- May 22nd 2008, 02:18 AMMoo
- May 22nd 2008, 02:21 AMIsomorphism
Ya obviously, We are taught limits first... how else will you learn the definition of differentiation?

Default

Quote:

Originally Posted by Isomorphism View Post

Are you introduced to differentiation? If not then you have to use limit definition of \lim_{n \to \infty} \left(1+\frac1{n}\right)^n = e

So follow what I said in post 2.

Is it possible that one is taught this limit before doing differentiation ?

Quote:

LOL ! I forgot... by the way, his/her previous thread was about implicit differentiation

- May 22nd 2008, 02:25 AMMoo
- May 22nd 2008, 02:42 AMGigabyte
Yea i have been introduced to differentiation. However i don't really understand that line. I tried to read about it but got more confused.

So from post 8 would you get????

lim as x tents to infinity

ln[(1+a/x)^(x/a)]^a = e^a - May 22nd 2008, 03:03 AMMoo
You just have to know that :

$\displaystyle \lim_{t \to 0} \frac{f(a+t)-f(a)}{t}=f'(a)$

Here, $\displaystyle a=1$ and $\displaystyle f(x)=\ln(1+t) \implies f'(x)=\frac{1}{1+t}$

Quote:

So from post 8 would you get????

lim as x tents to infinity

ln[(1+a/x)^(x/a)]^a = e^a

- May 22nd 2008, 04:10 AMGigabyte
wow that was a mammoth step. But i think i have caught up to you. Now i'm stuck again lol.

What is the derivative number, and how does it help to solve the next line of working in post 3?? Otherwise i think i've got it.

Thank you so much for your help so far. It's been awesome, i'm really learning alot. - May 22nd 2008, 05:56 AMMoo
The derivative number in x=a is actually the value of the derivative at this point x=a.

It's the same as taking the value of the derivative function in a.

This is the very first definition of derivative that you should have learnt :eek:

Is it all clear now ? (Worried)