$$\lim_{x \to \sqrt{3}^{-}} \sqrt{x^2-3}$$
What's the answer of this limit? There are two hyppothesis:
$0$ and undefined.
Undefined Because the square root of a negative number is undefined, $0$ because if we plug $\sqrt{3}$, we obtain $0$; I am not so sure. Please help
Edit:
Why do we calculate this limit?
According to my teacher, a function limit at sqrt(3) exists if limit at sqrt(3)- and limit at sqrt(3)+ both exists and are the same. So that's why we tried to find it.
Is that right?
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$\begingroup$A presumption of the usual definition of limit is that the object under consideration is a function. Note that $\sqrt{x^{2}-3}$ is not even defined if $-\sqrt{3} < x < \sqrt{3}$, so it is not legitimate to study the limit $\lim_{x \to \sqrt{3}-}\sqrt{x^{2}-3}$. However, the right-hand limit here exists and $=0$. What is the domain of the function $x \mapsto \sqrt{x^{2}-3}$?
$\endgroup$ 2 $\begingroup$What your teacher said is wrong.
Limits are computed inside the domain. For this function, both the ordinary and right-hand limits exist and are both $0$ (by plugging $\sqrt3$ and because the function is continuous). The left-hand limit is undefined, but this doesn't matter.
The situation would be different with a domain extending in the negatives with no left-hand limit, like
$$\begin{cases}x<0\to\sin\dfrac1x,\\x\ge0\to\sqrt x\end{cases}$$
(no left-hand and no ordinary limit but a right-hand limit)
or a domain extending in the negatives and with a different left-hand limit such as
$$\begin{cases}x<0\to-1,\\x\ge0\to\sqrt x\end{cases}$$
(both left-hand and right hand but no ordinary limit).
$\endgroup$ $\begingroup$The function $f (x) = \sqrt{x^2 - 3}$ is defined on $D = ({-\infty}, -\sqrt{3}] \cup [\sqrt{3}, {+\infty})$. To compute the limit of $f$ as $x$ approaches $\sqrt{3}$, you need to compute $f (x)$ for values of $x$ that are arbitrarily close to $\sqrt{3}$. And to compute $f (x)$ for any value $x$, you need $x$ to be in the domain of $f$.
So the limit $\lim_{x \to \sqrt{3}} f (x)$ has to be taken for $x > \sqrt{3}$, and will thus be $0$. This is an example where a right limit (approaching from strictly higher values) exists, but no left limit (approaching from strictly lower values) exist.
$\endgroup$ $\begingroup$There are a few things to consider:
0 because if we plug sqrt(3), we obtain 0
This is false. The value of a function at the limit point does not matter for the value of the limit. Plugging a value in is a trick that often works, but it only works if the function is continuous at the limit point (in your case only left continuous is necessary, since we have a limit from the left). Since the function is not defined left of $\sqrt{3}$, it is not left continuous at $\sqrt{3}$.
Undefined Because the square root of a negative number is undefined
The limit of undefined values is again undefined. This is true.
There is one more thing to consider: Complex numbers. The idea of complex numbers is that the square root of a negative number is defined. When we define $\sqrt{x^2-3}$ as a function from the real numbers to the complex numbers ($\mathbb{R} \to \mathbb{C}$) or as a function from the complex numbers to the complex numbers ($\mathbb{C} \to \mathbb{C}$), it is defined and continuous in $\sqrt{3}$. So this means plugging in $\sqrt{3}$ is allowed. So the limit is 0.
If you define $\sqrt{x^2-3}$ to be a function to complex numbers, then the limit is 0. This is why Wolfram Alpha says the limit is 0.
If you define $\sqrt{x^2-3}$ to be a function to real numbers, then the limit is undefined.
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