Value of an expression as two variables approach infinity I am currently reading the first edition of Abraham de Moivre's book: The Doctrine of Chances. I'm stuck at Problem V. I will cite the statement of the problem.

To find in how many Trials an Event will Probably Happen, or how many Trials will be requisite to make it indifferent to lay on its Happening or Failing; supposing that a is the number of chances for its Happening in any one Trial, and b the number of chances for its failing.

De Moivre does offer a solution for each of the problems in the book, however there are some details that are not treated rigurously. In his solution, he proposes for us to suppose that the proportion of $a$ to $b$ is as the proportion of $1$ to $q$. Therefore,  if $a=k$ then $b=kq$. Under this assumption —and being $x$ the number of trials—, the equation $(a+b)^x=2b^x$ becomes $(\frac{1}{q}+1)^x=2$. 
Afterwards, de Moivre proceeds to state what the value of $x$ must be according to a given value of $q$ in order to preserve the condition that $(\frac{1}{q}+1)^x=2$. 
If $q=1$ it's very easy to see that $x=1$ as well. 
However, it's not that easy to determine what the value of $x$ must be as $q$ approaches infinity. De Moivre merely mentions that in that case $x$ must also approach infinity, but he offers no proof of this. My question is how can this be proved. I've tried to do it by calculating $lim_{q\to\infty}(\frac{1}{q}+1)^x$ and then thinking about what happens when the value of $x$ starts growing. 
So, first I use Newton's binomial theorem to obtain the following expresion. $$\big(\frac{1}{q}+1\big)^x=\big(\frac{1}{q}\big)^x+x\cdot\big(\frac{1}{q}\big)^{x-1}+\frac{x(x-1)}{2!}\cdot\big(\frac{1}{q}\big)^{x-2}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\frac{x(x-1)(x-2)}{3!}\cdot\big(\frac{1}{q}\big)^{x-3}+...+x\cdot\big(\frac{1}{q}\big)+1$$ Since each binomial coefficient appears two times —except the coefficient of the middle if $x$ is not an even number—, I proceed to factorize.
$$\big(\frac{1}{q}+1\big)^x=\Big[\big(\frac{1}{q}\big)^x+1\Big]+x\Big[\big(\frac{1}{q}\big)^{x-1}+\frac{1}{q}\Big]+\frac{x(x-1)}{2!}\Big[\big(\frac{1}{q}\big)^{x-2}+\big(\frac{1}{q}\big)^2\Big]+\frac{x(x-1)(x-2)}{3!}\Big[\big(\frac{1}{q}\big)^{x-3}+\big(\frac{1}{q}\big)^3\Big]+...$$
Next, I take the limit.
$$lim_{q\to\infty}\big(\frac{1}{q}+1\big)^x=1+lim_{q\to\infty}\big(\frac{1}{q}\big)^x+x\cdot\lim_{q\to\infty}\Big[\frac{1+q^{2-x}}{q}\Big]+\frac{x(x-1)}{2!}\lim_{q\to\infty}\Big[\frac{1+q^{4-x}}{q^2}\Big]+\frac{x(x-1)(x-2)}{3!}\lim_{q\to\infty}\Big[\frac{1+q^{6-x}}{q^3}\Big]+...$$
By using L'Hôpital's rule we get the following.
$$lim_{q\to\infty}\big(\frac{1}{q}+1\big)^x=1+lim_{q\to\infty}\big(\frac{1}{q}\big)^x+x\cdot\lim_{q\to\infty}\Big[(2-x)q^{1-x}\Big]+\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{x(x-1)}{2!}\lim_{q\to\infty}\Big[\frac{(4-x)(3-x)q^{2-x}}{2}\Big]+\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{x(x-1)(x-2)}{3!}\lim_{q\to\infty}\Big[\frac{(6-x)(5-x)(4-x)q^{3-x}}{6}\Big]+...$$
Having reached this point, it seems to me that if we made $x$ approach infinity then all the limits at the right-hand side of the equation would be equal to $0$. My reasoning is this. Take, for example, the expression $\lim_{q\to\infty}\Big[\frac{(6-x)(5-x)(4-x)q^{3-x}}{6}\Big]$. As $x$ started growing in value, which function would grow faster? $(6-x)(5-x)(4-x)$ or $q^{x-3}$? It all comes down to calculating the derivatives. The derivative of $(6-x)(5-x)(4-x)$ is $-3x^2+30x-74$, while the derivative of $q^{x-3}$ is $(x-3)q^{x-4}$. As soon as $x$ reaches the value of $5$, $q^{x-3}$ starts growing infinite times more than $(6-x)(5-x)(4-x)$ —remember that $q$ approaches infinity. And the former function will continue growing much more rapidly than the latter. 
Therefore, contrary to what de Moivre says, I conclude that $lim_{q\to\infty}\big(\frac{1}{q}+1\big)^x=1$ as $x$ approaches infinity. Surely, I must have made a mistake but I can't identify it.

$e^{\ln 2}=2$. $\lim_{q\to \infty}(1+\frac{1}{q})^q=e$  Therefore $\lim_{q\to \infty}(1+\frac{1}{q})^x=2$, when $x=q\ln 2$.
