Nasty limit...

**centenial** · May 14th 2011, 09:13 PM

I have this really nasty limit that I've been banging my head on all night. The question asks me to calculate the following:

$\lim_{n\to\infty}\frac{n \times \left(\frac{2n-2}{2n-1}\right)^{2n-2}}{2n-1}$

The limit evaluates to $\frac{\infty}{\infty}$ , so my first instinct was to use L'Hospital's Rule and take the limit of $\frac{f'(x)}{g'(x)}$ . But that didn't really simplify it; doing that just gave another nasty limit. Is there a simpler approach?

According to Wolfram Alpha, the limit is equal to $\frac{1}{2e}$ . But it just says "series expansion at $n=\infty$ " without showing any of the steps.

I know that $e$ is the sum of an infinite series. But how do you recognize that from the limit? And how do you get to $\frac{1}{2e}$ ?

Any pointers would be greatly appreciated!

- A really puzzled math student.

**FernandoRevilla** · May 14th 2011, 09:27 PM

Express

$L=\lim_{n\to +\infty}\dfrac{n}{2n-1}\cdot \left(1-\dfrac{1}{2n-1}\right)^{2n-2}$

and apply the definition of the number e .

**TheCoffeeMachine** · May 14th 2011, 09:32 PM

Write the expression under the limit (I wonder what it's called?) as:

$\left( \frac{n}{2n-1}\right)\left(\frac{2n-1}{2n-2}\right)\left(1-\frac{1}{2n-1}\right)^{2n-1}$

EDIT: far too late! O_o

**centenial** · May 14th 2011, 09:43 PM

You guys are amazing. Thanks so much!

**Prove It** · May 14th 2011, 09:44 PM

$\displaystyle \begin{align*}\lim_{n \to \infty}\frac{n\left(\frac{2n - 2}{2n - 1}\right)^{2n-2}}{2n - 1} &= \lim_{n \to \infty}\left(\frac{n}{2n - 1}\right) \lim_{n \to \infty}\left(\frac{2n - 2}{2n - 1}\right)^{2n - 2}\\ &= \frac{1}{2}\lim_{n \to \infty}\left(\frac{2n - 2}{2n - 1}\right)^{2n - 2} \\ &= \frac{1}{2}\lim_{n \to \infty}e^{\ln{\left[\left(\frac{2n-2}{2n - 1}\right)^{2n-2}\right]}} \\&= \frac{1}{2}e^{\lim_{n \to \infty}\ln{\left[\left(\frac{2n-2}{2n-1}\right)^{2n-2}\right]}}\\ &= \frac{1}{2}e^{\lim_{n \to \infty}\left[(2n - 2)\ln{\left(\frac{2n - 2}{2n - 1}\right)}\right]} \\ &= \frac{1}{2}e^{\lim_{n \to \infty}\left[\frac{\ln{(2n - 2)} - \ln{(2n - 1)}}{(2n - 2)^{-1}}\right]}\\ &= \frac{1}{2}e^{\lim_{n \to \infty}\left[\frac{\frac{1}{n - 1} - \frac{2}{2n - 1}}{-2(2n - 2)^{-2}}\right]}\,\,\textrm{by L'Hospital's Rule}\end{align*}$

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