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Wikkibahi · Dec 1, 2011, 11:19 PM

I attach the Pdf Math File which is shown as below. Please Open it and Solve my Problem...?
Please tell me as soon as possible.
Thanks a lot for this....1

Aurora2000 · Dec 2, 2011, 05:53 AM

It is the same, x=-(t+1) or x=-t. Personally I would have put x=-t.

ebaines · Dec 2, 2011, 07:04 AM

If you replace x with -t you get:

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

Not sure how to proceed with that. But if you replace x with -(t+1) you get:

$ \lim_{t \to \infty} (1 - \frac 1 {t+1} )^{-(t+1)} = \lim_{t \to \infty} (\frac t {t+1} )^{-(t+1)} = \lim_{t \to \infty} (\frac {t+1} t)^{(t+1)}= \lim_{t \to \infty} (1 + \frac 1 t)^{(t+1)} $

which was already shown to equal e.

Aurora2000 · Dec 2, 2011, 07:35 AM

With x=-t you get

$\lim_{t\rightarrow +\infty} (\frac{t}{t-1})^t = \lim_{t\rightarrow +\infty} (1+\frac{1}{t-1})^t= \lim_{t\rightarrow +\infty} ((1+\frac{1}{t-1})^{t-1})^{\frac{t}{t-1}}$