Limit with terms very similar to those that should give an exponential function

Question

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?

Answer 1

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

Answer 2

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?)

Answer 3

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}