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.
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: $$\lim_{t \to +\infty} -\frac{1 + t}{1 + tx} = \lim_{x \to +\infty} -\frac{1 + \frac 1t}{x + \frac 1t} = -\frac1x.$$
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
$$\frac{1-e^{-1}}{1-e^{1/(1+t)}}\to -\infty$$
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,
$$-\frac1x\frac{e^{0/x}}{e^0}.$$
