How to prove that $\lim_{x \to \infty} (\sqrt{x + \gamma} / x) = \lim_{x \to \infty} 1 / \sqrt{x} = 0$?

Question

As the title specifies:

How do I (rigorously) prove that

$$\lim_{x \to \infty} \frac{\sqrt{x + \gamma}}{x} = \lim_{x \to \infty} \frac{1}{\sqrt{x}} = 0?$$

($\gamma \in \mathbb{R}.$)

Answer 1

Suppose $x>|\gamma|$ (i.e $x$ is sufficiently positive to make the manipulations below valid). Then, \begin{align} \frac{\sqrt{x+\gamma}}{x}&=\frac{1}{\sqrt{x}}\cdot\sqrt{1+\frac{\gamma}{x}}. \end{align} Since this equality holds for all $x\in\Bbb{R}$ such that $|x|>\gamma$, we can consider the question of taking the limit as $x\to\infty$. Now,

• What is the limit of $\frac{1}{x}$ as $x\to\infty$?
• What is the limit of the first term, $\frac{1}{\sqrt{x}}$?
• What is the limit of the stuff inside the square root, $1+\frac{\gamma}{x}$, for the second term? So including the square root on the second term, what’s the limit?

In other words, by the rules for limits (product and composition), the limit is___?

So, you just have to prove the basic theorems and apply them carefully. Of course, this particular example is simple enough that you could combine all the steps into a single $\epsilon$-$N$ proof, but this is highly inefficient.