Solving restricted/one-sided limits

Question

Is the function $f(x) = \sqrt{x^{2} - 9}$ continuous at $x = 3$?

I tried to check by solving each requirement for continuity, which are the following:

1. The limit at the point must exist;
2. The function at the point must exist;
3. The limit at the point must be equal to the function at the point.

The second requirement can be easily checked as 'satisfied' as $f(3) = 0$. Thus, I only need to check if the first requirement is satisfied.

I started by solving for the left-hand limit as follows: \[L^{-} = \lim_{x \to 3^{-}}\left[\sqrt{x^{2} - 9}\right]\]

Then, I eventually got $L^{-}$ to be zero.

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My question is this: Is it correct to rewrite the problem by substituting $x = 3 - a$, where $a \geqslant 0$, then solving for the limit as $a$ approaches zero?

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Edit: This is a question about solving one-sided limits, and not about the continuity of the square root function.

Answer 1

You do not need to evaluate for left-hand limit simply because it does not belong to the domain of the function. In general for obtaining continuity of boundary point we consider the limiting value of the function as the value of the one-sided limit. Thus if,

$$\lim_{x\rightarrow 3^+}f(x)=f(3)$$

you can say that the function is continuous at $x=3$. as the function is defined only for $|x|\ge3$ assuming you are working over $\mathbb R$.

Answer 2

About substitutions in limits in general:

In general, that's absolutely right (here, in particular, something more important is to be noted. See the Addendum). $x\to 3^-$ can be thought of as $x = 3-h$ where $h\to 0^+$ or $x = 3+h$ where $h\to 0^-$. Guess what, that's exactly the substitution you made.

Subtle point to note: You say $x = 3-a$, where $a\ge 0$. While that is perfectly fine (in general it is fine, but for this particular case please check the addendum), I just want to mention that $a > 0$ is no problem either. When talking about $\lim\limits_{a\to 0^+}f(a)$ for some $f$, $a$ is approaching $0$ from the positive side, but is never equal to $0$. Hence, $a>0$ suffices and the substitution $x = 3-a$ works.

Before making substitutions though, one must be extremely careful about the domain of the function.

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Addendum/Warning:
The domain of $f(x) = \sqrt{x^2-9}$ is $x^2 \ge 9$, i.e. $|x| \ge 3$ or $x \in (-\infty,-3]\cup [3,\infty)$. So, it does not make sense to talk about the left hand limit at $3$, since values immediately to the left of $3$ are not in the domain of $f$. You may ask, does $$\lim_{x\to 3} \sqrt{x^2-9}$$ still exist? Yes, it does! This is because the only one-sided limit we can even talk about at $3$ is the right-handed one. We define the limit at $3$ to be equal to this one-sided limit.