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I have been trying to solve the following limit but am completely stuck.

$$\lim_{\alpha \rightarrow \infty} 1-\left( \frac{y+\alpha}{\alpha-1} \right)^{-\alpha}$$

I have tried inverting the ratio and came up with the following expression:

$$ 1 - \lim_{\alpha \rightarrow \infty} \left( 1-\frac{y+1}{y+\alpha}\right)^\alpha$$

Which roughly resembles the exponential function:

$$\lim_{\alpha \rightarrow \infty} \left( 1- \frac{x}{\alpha} \right)^\alpha = \exp(-x)$$

Except for the additive term in the denominator. Is there a u-substitution type trick to this?

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$$\lim_{\alpha \rightarrow \infty} \left( 1-\frac{y+1}{y+\alpha}\right)^\alpha=\lim_{\alpha \rightarrow \infty} \dfrac{\left( 1-\frac{y+1}{y+\alpha}\right)^{y+\alpha}}{\left( 1-\frac{y+1}{y+\alpha}\right)^{y}}=\exp(-(y+1))$$

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  • $\begingroup$ I see. I hadn't accounted for application of L'Hospital's Rule in this case. The denominator effectively converges to 1 and that makes the numerator gives us the desired quantity. Thank you! $\endgroup$ – AdamO Sep 29 '14 at 4:42
  • $\begingroup$ You're welcome, @ashkan! $\endgroup$ – Zircht Sep 29 '14 at 4:43
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If you're already a little advanced in functions and their limits, you could try

$$\lim_{\alpha\to\infty}\frac{\log\left(1-\frac{y+1}{y+\alpha}\right)}{\frac1\alpha}\stackrel{\text{l'Hospital}}=\lim_{\alpha\to\infty}\frac{\frac{y+\alpha}{\alpha-1}\cdot\frac{y+1}{(y+\alpha)^2}}{-\frac1{\alpha^2}}=-\lim_{\alpha\to\infty}\frac{\alpha^2}{\alpha-1}\frac{y+1}{\alpha+y}=-(y+1)$$

So the original limit is $\;e^{-(y+1)}\;$ (why?)

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  • $\begingroup$ This is a clever answer. What, in general, are sufficient conditions so that limits can be "passed through" functions such as exponents and logarithms? Analytic? Smooth? $\endgroup$ – AdamO Sep 29 '14 at 4:38
  • $\begingroup$ Hi ashkan, about the comment above I believe that the most general condition for one to be able to pass a limit through a function is that the function must be continuous. For example $\lim_{x \to \infty} cos(1/x) = cos(\lim_{x \to \infty}(1/x)) =1$. The standard proof has to do with the definition of a limit using sequences. $\endgroup$ – Rogelio Molina Oct 1 '14 at 0:46
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You can try the $u =y + \alpha$ substitution, in this way you don't need L'Hopital's rule:

\begin{eqnarray} \lim_{\alpha \to \infty} \left( 1- \frac{1+y}{y+\alpha} \right)^{\alpha} = \lim_{u \to \infty} \left( 1- \frac{1+y}{u} \right)^{u-y} = \lim_{u \to \infty} \left( 1- \frac{1+y}{u} \right)^{u}/ \lim_{u\to \infty} \left(1 -\frac{1+y}{u} \right)^y = e^{-(1+y)}/1 = e^{-(1+y)} \end{eqnarray}

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  • $\begingroup$ I see you break the exponent $u-y$ into a ratio and apply the limit to either term. Is that not L'Hospital's Rule? $\endgroup$ – AdamO Sep 29 '14 at 4:43
  • $\begingroup$ No, L'Hospital's rule would require computing derivatives, that is the way in which Timbuc above has done it. What is being used here is the theorem that the limit of a quotient is equal to the quotient of the limits, provided each limit in the quotient exists separately. $\endgroup$ – Rogelio Molina Sep 30 '14 at 0:36
  • $\begingroup$ You are right Rogelio. For whatever reason this had eluded me. $\endgroup$ – AdamO Sep 30 '14 at 1:17

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