Consider a linear function $f(x) = ax+1$ which is defined on all $x \neq -1/a$ for some fixed constant $a$ . Given integer $n\geq 1$, I would like to evaluate the following limit $$ \lim_{x \to -1/a} \bigg[ \left(1+{ \sqrt{ f(x) } }\right)^{n} \bigg(1+ \frac{x}{\sqrt{f(x)}}\bigg) + \left(1-{ \sqrt{ f(x) } }\right)^{n} \bigg(1- \frac{x}{\sqrt{f(x)}}\bigg) \bigg]. $$
My attempt: First note that $\lim_{x\to -1/a} f(x) = 0$. Then I write the limit above as a sum of two limits $$ \lim_{x \to -1/a} F_1(x) + \lim_{x \to -1/a} F_2(x) $$ where $F_1(x) = \left(1+{ \sqrt{ f(x) } }\right)^{n} \bigg(1+ \frac{x}{\sqrt{f(x)}}\bigg)$ and $F_2(x) = \left(1-{ \sqrt{ f(x) } }\right)^{n} \bigg(1- \frac{x}{\sqrt{f(x)}}\bigg)$. Then for evaluating the first term $\lim_{x \to -1/a} F_1(x) $, use product rule for limits, I obtain $$ \lim_{x \to -1/a} \left(1 + { \sqrt{ f(x) } }\right)^{n} \lim_{x \to -1/a} \bigg(1+ \frac{x + 1}{\sqrt{f(x)}}\bigg) = 1 \cdot \lim_{x \to -1/a} \bigg(1+ \frac{x + 1}{\sqrt{f(x)}}\bigg) = 1 + \frac{-1/a+1}{"0"} $$ but the second limit blows up to infinity (in fact, as seen in the last term, it is not in the standard indeterminate form such as 0/0 so the L'hosptal's rule cannot apply here). Same phenomona happens when I evaluating $\lim_{x \to -1/a} F_2(x) $ and not sure what to do with it. I would appreicate any help or comment. Thanks.
Your problem, which is a common one, is trying to to add divergent limits. You have to be sure limits converge before you add them.
If your objection to l'Hôpital's rule is practical rather than moral, you should note that it can often be made to apply to $\infty - \infty$ by use of the exponential. Regardless, in this case it's no use. You can't write it as the sum of two limits because both those limits diverge. I'll assume $a \neq 0$ even though you don't say it. Let's first try regrouping terms.
$$\lim_{x \to -1/a} \bigg[ \left(1+{ \sqrt{ f(x) } }\right)^{n} \bigg(1+ \frac{x}{\sqrt{f(x)}}\bigg) + \left(1-{ \sqrt{ f(x) } }\right)^{n} \bigg(1- \frac{x}{\sqrt{f(x)}}\bigg) \bigg] = \lim_{x \to -1/a} \bigg[ 2\left(1+{ \sqrt{ f(x) } }\right)^{n} + \left(1+{ \sqrt{ f(x) } }\right)^{n} \bigg(\frac{x}{\sqrt{f(x)}}\bigg) -\left(1-{ \sqrt{ f(x) } }\right)^{n} \bigg(\frac{x}{\sqrt{f(x)}}\bigg) \bigg] = 2 + \lim_{x \to -1/a} \left(\frac{x}{\sqrt{f(x)}}\left((1+\sqrt{f(x)}\right)^n-\left(1-\sqrt{f(x)}\right)^n\right)$$
From which the binomial expansion yields $2 - \frac{2n}{a}$. Note that wherever limits are added or multiplied, it's ultimately shown both converge.