Finding the limit of ((x-1)/x)^x as x->infinity

• August 10th 2009, 07:21 AM
sevenquid
Finding the limit of ((x-1)/x)^x as x->infinity
Dear all,
I am trying to find, analytically, the limit of ((x-1)/x)^x as x->infinity.

I tried l'Hopital's rule but it seems to make it more complicated. I know that the limit is 1/e but need to show my working and understand why it is 1/e.

Hope you can help!
Many thanks
Mike
• August 10th 2009, 07:43 AM
chengbin
We know $\lim_{x \to 0} (1+x)^\frac {1}{x}=e$

Let's simplify this first.

$\lim_{x \to \infty} \bigg ( \frac {x-1}{x}\bigg )^x$ = $\lim_{x \to \infty} \bigg ( 1-\frac{1}{x}\bigg )^x$

Let $-\frac {1}{x}=h$, as $x \to \infty, h \to 0$

$\therefore \lim_{x \to \infty} \bigg ( \frac {x-1}{x}\bigg )^x=\lim_{h \to 0} (1+h)^{-\frac {1}{h}}=\frac {1}{e}$
• August 10th 2009, 07:45 AM
Plato
Take note that $\left( {\frac{{x - 1}}{x}} \right)^x = \left( {1 + \frac{{ - 1}}{x}} \right)^x$.
That is a standard $e\text{-form}$.
• August 10th 2009, 07:47 AM
Random Variable
$\lim_{x \to \infty} \ln \Big(\frac{x-1}{x}\Big)^{x}$

$= \lim_{x \to \infty} x \ln \Big(\frac{x-1}{x}\Big)$

$= \lim_{x \to \infty} \frac{\ln \Big(\frac{x-1}{x}\Big)}{\frac{1}{x}}$

$= \lim_{x \to \infty} \frac{\frac{x}{x-1} \frac{1}{x^{2}}}{\frac{-1}{x^{2}}}$

$= \lim_{x \to \infty} \frac{-x}{x-1}$

$= \lim_{x \to \infty} \frac{-1}{1-\frac{-1}{x}} = -1$

so $\lim_{x \to \infty} \Big(\frac{x-1}{x}\Big)^{x} = e^{-1}= \frac{1}{e}$
• August 10th 2009, 07:52 AM
skeeter
Quote:

Originally Posted by sevenquid
Dear all,
I am trying to find, analytically, the limit of ((x-1)/x)^x as x->infinity.

I tried l'Hopital's rule but it seems to make it more complicated. I know that the limit is 1/e but need to show my working and understand why it is 1/e.

Hope you can help!
Many thanks
Mike

let $y = \left(\frac{x-1}{x}\right)^x$

$\ln{y} = x \cdot \ln\left(\frac{x-1}{x}\right) = \frac{\ln\left(\frac{x-1}{x}\right)}{\frac{1}{x}} = \frac{\ln\left(1 - \frac{1}{x}\right)}{\frac{1}{x}}$

$\lim_{x \to \infty} \frac{\ln\left(1 - \frac{1}{x}\right)}{\frac{1}{x}}$

use L'Hopital ...

$\lim_{x \to \infty} \frac{\frac{\frac{1}{x^2}}{1 - \frac{1}{x}}}{-\frac{1}{x^2}}$

$\lim_{x \to \infty} \frac{-1}{1 - \frac{1}{x}} = -1$

as $x \to \infty$ , $\ln{y} \to -1$ ... $y \to e^{-1}
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