Evaluating a limit at infinity

Question

I was doing a physics problem when I had to compute an integral of the form $$I=\int\frac{dx}{\sqrt{a+x^2}} $$ for $a>0$, which is easily eavluated by making the substitution $x=\sqrt{a}\tan^{-1}(\theta)$ to yield $$I=\frac{1}{a}\sin\big(\tan^{-1}\big(\frac{u}{\sqrt{a}}\big)\big)+C=\frac{1}{a}\cdot\frac{x}{\sqrt{a+x^2}}+C$$However, I needed to evaluate $$\lim_{x\to\infty} \frac{x}{\sqrt{a+x^2}}-\lim_{x\to-\infty}\frac{x}{\sqrt{a+x^2}}$$ in the process. I manipulated the fraction by noticing that $$\frac{x}{\sqrt{a+x^2}}=\frac{1}{\sqrt{\frac{a}{x^2}+1}}$$ and as $x\to\infty$, it is clear that the limit goes to $1$. But the limit as $x\to-\infty$ is less clear. I tried to make the substitution $x\mapsto-y$ for $y>0$, which seemed to make more sense, as we obtain $$\lim_{x\to-\infty}\frac{x}{\sqrt{a+x^2}}=-\lim_{y\to\infty}\frac{y}{\sqrt{a+y^2}}=-1$$ but I am not sure how valid that method is. Is there a better way to evaluate the limit as $x\to-\infty$?

Answer 1

What you did is correct. One simple way to see this is to observe that as $x$ becomes arbitrarily large, the contribution of $a$ to the limit is negligible... In other words, this is the idea behind the comparison test. Anyway, the point is that you can pretend the $a$ is not there... Now with the understanding that $\sqrt{}$ means positive square root, you can see why what you did is correct...