# Taking a Limit for a Product of Functions Involving Quotient

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.

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.