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Given the following fraction: $$\frac{1-\exp\left(-\frac{1}{1+tx}\right)}{1-\exp\left(\frac{1}{1+t}\right)}$$

I need to find the limit as $t$ tends to infinity, so:

$$\lim_{t\rightarrow\infty} \frac{1-\exp\left(-\frac{1}{1+tx}\right)}{1-\exp\left(\frac{1}{1+t}\right)}$$

A formal proof is not needed.

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  • $\begingroup$ For $u$ close to $0$, $\exp(-u)$ is approximately $1-u$. $\endgroup$ – John Dawkins Jan 2 '16 at 15:57
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We see that for $t\to \infty$ the arguments of the exponential functions tend to zero. Therefore, since

$$1+x\le e^x\le \frac{1}{1-x}=1+x+O(x^2)$$

for $x<1$, we can simply write heuristically

$$e^{x}\approx 1+x$$

Then, we have for $x\ne 0$

$$\frac{1-e^{-1/(1+tx)}}{1-e^{1/(1+t)}}\approx -\frac{1+t}{1+tx}\to -\frac1x$$

And for $x=0$ we have

$$\frac{1-e^{-1}}{1-e^{1/(1+t)}}\to -\infty$$

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  • $\begingroup$ Happy New Year !! my friend. $\endgroup$ – Claude Leibovici Jan 2 '16 at 16:59
  • $\begingroup$ @ClaudeLeibovici Happy New Year to you also!! I hope you are doing well. $\endgroup$ – Mark Viola Jan 2 '16 at 17:02
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The function $f(x) = 1 - \exp(x)$ can be approximated, around $x = 0$, by the function $-x$. You can graph them to check this or just use the McLaurin expansion of $\exp x$.

The limit is then: $$\lim_{t \to +\infty} -\frac{1 + t}{1 + tx} = \lim_{x \to +\infty} -\frac{1 + \frac 1t}{x + \frac 1t} = -\frac1x.$$

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When $t$ grows, the terms $1$ in the exponentials become neglectible and one can trade the limit at infinity for one at zero:

$$\lim_{h\to0}\frac{1-e^{-h/x}}{1-e^h}.$$

Then by L'Hospital,

$$-\frac1x\frac{e^{0/x}}{e^0}.$$

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