please help
evaluate lim as x goes to infinity (1+(a/x))^x
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 $\displaystyle \left(1+\frac ax\right)^x=e^{\color{red}\ln\left[\left(1+\frac ax\right)^x\right]}$
---> $\displaystyle \boxed{\lim_{x \to \infty} \left(1+\frac ax\right)^x=e^a}$
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.
$\displaystyle \ln(1)=0$, so I could add it, right ?
YesWhen you take (a) out the front of the limit. You take (a) out the front cos it's a constant right?
Have look here : Derivative - Wikipedia, the free encyclopediaThen 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.
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 ?
Oh ok thenLOL ! I forgot... by the way, his/her previous thread was about implicit differentiation
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 $\displaystyle f(x)=\ln(1+t) \implies f'(x)=\frac{1}{1+t}$
yeah, this is the resultSo from post 8 would you get????
lim as x tents to infinity
ln[(1+a/x)^(x/a)]^a = e^a
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.
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
Is it all clear now ?